1.1.1

Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure.

x1+4x2	=	5
4x1+5x2	=	−13

Find the solution to the system of equations.

(−7,3)

(Simplify your answer. Type an ordered pair.)

1.1.1

Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure.

x1+5x2	=	9
4x1+5x2	=	−9

Find the solution to the system of equations.

(−6,3)

(Simplify your answer. Type an ordered pair.)

1.1.2

Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure.

5x1+10x2	=	20
4x1+7x2	=	18

Find the solution to the system of equations.

(8,−2)

(Simplify your answer. Type an ordered pair.)

1.1.2

Solve the system by using elementary row operations on the equations. Follow the systematic elimination procedure.

6x1+12x2	=	18
4x1+7x2	=	15

Find the solution to the system of equations.

(9,−3)

(Simplify your answer. Type an ordered pair.)

1.1.3

Find the point x1,x2 that lies on the line x1+2x2=5 and on the line x1−x2=−1. See the figure.

x1 x2

The point

x1,x2

that lies on the line

x1+2x2=5

and on the line

x1−x2=−1

is

(1,2).

(Simplify your answer. Type an ordered pair.)

1.1.3

Find the point x1,x2 that lies on the line x1+2x2=10 and on the line x1−x2=1. See the figure.

x1 x2

The point

x1,x2

that lies on the line

x1+2x2=10

and on the line

x1−x2=1

is

(4,3).

(Simplify your answer. Type an ordered pair.)

1.1.13

Solve the system.

x1			−	3x3	=	7
4x1	+	4x2	+	x3	=	23
		2x2	+	3x3	=	1

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The unique solution of the system is

4,2,−1.

(Type integers or simplified fractions.)

The system has infinitely many solutions.

The system has no solution.

1.1.13

Solve the system.

x1			−	6x3	=	22
2x1	+	4x2	+	x3	=	21
		2x2	+	3x3	=	−1

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The unique solution of the system is

4,4,−3.

(Type integers or simplified fractions.)

The system has infinitely many solutions.

The system has no solution.

1.1.26

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

6	4	h
−3	−2	5

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrix is the augmented matrix of a consistent linear system if

h=−10.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system if

h≠.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system for every value of h.

The matrix is not the augmented matrix of a consistent linear system for any value of h.

1.1.26

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

6	4	h
−3	−2	3

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrix is the augmented matrix of a consistent linear system if

h=−6.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system if

h≠.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system for every value of h.

The matrix is not the augmented matrix of a consistent linear system for any value of h.

1.1.28

Determine whether the statement below is true or false. Justify the answer.

Elementary row operations on an augmented matrix never change the solution set of the associated linear system.

Choose the correct answer below.

The statement is true. Elementary row operations are always applied to an augmented matrix after the solution has been found.

The statement is false. Interchanging two rows never changes the solution set of the associated linear system. However, replacing one row by the sum of itself and a multiple of another row can change the solution set of that system.

The statement is true. Each elementary row operation replaces a system with an equivalent system.

The statement is false. Interchanging two rows never changes the solution set of the associated linear system. However, scaling a row by a nonzero constant can change the solution set of that system.

1.1.31

Determine whether the statement below is true or false. Justify the answer.

The solution set of a linear system involving variables

x1,

...,

xn

is a list of numbers

s1, ..., sn

that makes each equation in the system a true statement when the values

s1,

...,

sn

are substituted for

x1,

...,

xn,

respectively.

Choose the correct answer below.

The statement is true. It is the definition of the solution set of a linear system.

The statement is true. The solution set of a linear system will have the same number of elements as the list of the variables in the system.

The statement is false. The given description is of a single solution of such a system. The solution set of the system consists of all possible solutions.

The statement is false. The solution set of such a system consists of all possible orderings of the numbers

s1,

...,

sn.

1.1.31

Determine whether the statement below is true or false. Justify the answer.

The solution set of a linear system involving variables

x1,

...,

xn

is a list of numbers

s1, ..., sn

that makes each equation in the system a true statement when the values

s1,

...,

sn

are substituted for

x1,

...,

xn,

respectively.

Choose the correct answer below.

The statement is true. The solution set of a linear system will have the same number of elements as the list of the variables in the system.

The statement is false. The solution set of such a system consists of all possible orderings of the numbers

s1,

...,

sn.

The statement is false. The given description is of a single solution of such a system. The solution set of the system consists of all possible solutions.

The statement is true. It is the definition of the solution set of a linear system.

1.2.2

Determine which matrices are in reduced echelon form and which others are only in echelon form.

a. 1 0 0 0 0 1 0 0 0 0 1 1

b. 1 0 0 0 0 5 0 0 0 0 1 1

c. 1 2 0 0 0 0 0 0 0 0 1 0 0 0 0 1

Is matrix a in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix b in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

Is matrix c in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

1.2.2

Determine which matrices are in reduced echelon form and which others are only in echelon form.

a. 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0

b. 1 0 0 0 0 1 0 0 0 0 1 1

c. 1 5 0 0 0 0 0 0 0 0 1 0 0 0 0 1

Is matrix a in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix b in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix c in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

1.2.2

Determine which matrices are in reduced echelon form and which others are only in echelon form.

a. 1 0 0 0 0 1 0 0 0 0 1 1

b. 1 1 0 1 1 0 6 0 6 6 0 0 0 3 3 0 0 0 0 2

c. 1 6 0 0 0 0 0 0 0 0 1 0 0 0 0 1

Is matrix a in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix b in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

Is matrix c in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

1.2.2

Determine which matrices are in reduced echelon form and which others are only in echelon form.

a. 1 1 0 1 1 0 6 0 6 6 0 0 0 4 4 0 0 0 0 2

b. 1 0 1 1 0 1 1 1 0 0 0 0

c. 1 6 0 0 0 0 0 0 0 0 1 0 0 0 0 1

Is matrix a in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix b in reduced echelon form, echelon form only, or neither?

echelon form only

reduced echelon form

neither

Is matrix c in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

1.2.2

Determine which matrices are in reduced echelon form and which others are only in echelon form.

a. 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0

b. 1 0 1 1 0 1 1 1 0 0 0 0

c. 0 0 0 0 1 4 0 0 0 0 1 0 0 0 0 1

Is matrix a in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

Is matrix b in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

Is matrix c in reduced echelon form, echelon form only, or neither?

reduced echelon form

echelon form only

neither

1.2.5

Describe the possible echelon forms of a nonzero

2×2

matrix.

Select all that apply. (Note that leading entries marked with a

may have any nonzero value and starred entries

(*)

may have any value including zero.)

1.2.5

Describe the possible echelon forms of a nonzero

2×2

matrix.

Select all that apply. (Note that leading entries marked with a

may have any nonzero value and starred entries

(*)

may have any value including zero.)

1.2.6

Describe the possible echelon forms of a nonzero

6×2

matrix.

Select all that apply. (Note that leading entries marked with a

may have any nonzero value and starred entries

(*)

may have any value including zero.)

1.2.6

Describe the possible echelon forms of a nonzero

6×2

matrix.

Select all that apply. (Note that leading entries marked with a

may have any nonzero value and starred entries

(*)

may have any value including zero.)

1.2.7

Find the general solution of the system whose augmented matrix is given below.

	1	3	4	8
	2	6	5	7

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

x1=−4−3x2

x2 is free

x3=3

x1=

x2 is free

x3 is free

x1=

x2=

x3=

The system has no solution.

1.2.7

Find the general solution of the system whose augmented matrix is given below.

	1	4	3	3
	2	8	1	−4

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

x1=

x2 is free

x3 is free

x1=

x2=

x3=

x1=−3−4x2

x2 is free

x3=2

The system has no solution.

1.2.8

Find the general solution of the system whose augmented matrix is given below.

	1	4	0	9
	2	7	0	15

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

x1=−3

x2=3

x3 is free

x1=

x2 is free

x3 is free

x1=

x2=

x3=

The system has no solution.

1.2.8

Find the general solution of the system whose augmented matrix is given below.

	1	1	0	−4
	3	2	0	−15

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

x1=−7

x2=3

x3 is free

x1=

x2=

x3=

x1=

x2 is free

x3 is free

The system has no solution.

1.2.9

Find the general solution of the system whose augmented matrix is given below.

	0	1	−4	6
	1	−4	14	−21

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=

x2 is free

x3 is free

x1=

x2=

x3=

x1=3+2x3

x2=6+4x3

x3 is free

The system has no solution.

1.2.9

Find the general solution of the system whose augmented matrix is given below.

	0	1	−2	3
	1	−3	1	−5

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=4+5x3

x2=3+2x3

x3 is free

x1=

x2=

x3=

x1=

x2 is free

x3 is free

The system has no solution.

1.2.11

Find the general solution of the system whose augmented matrix is given below.

3	−2	7	0
12	−8	28	0
9	−6	21	0

Choose the correct answer below.

x1=23x2−73x3

x2 is free

x3 is free

x1=−3x2

x2=2x3

x3 is free

x1=3

x2=−2

x3=7

The system has no solutions.

1.2.11

Find the general solution of the system whose augmented matrix is given below.

5	−3	7	0
10	−6	14	0
15	−9	21	0

Choose the correct answer below.

x1=35x2−75x3

x2 is free

x3 is free

x1=−5x2

x2=3x3

x3 is free

x1=5

x2=−3

x3=7

The system has no solutions.

1.2.11

Find the general solution of the system whose augmented matrix is given below.

2	−3	9	0
4	−6	18	0
6	−9	27	0

Choose the correct answer below.

x1=32x2−92x3

x2 is free

x3 is free

x1=−2x2

x2=3x3

x3 is free

x1=2

x2=−3

x3=9

The system has no solutions.

1.2.11

Find the general solution of the system whose augmented matrix is given below.

3	−2	4	0
9	−6	12	0
6	−4	8	0

Choose the correct answer below.

x1=23x2−43x3

x2 is free

x3 is free

x1=−3x2

x2=2x3

x3 is free

x1=3

x2=−2

x3=4

The system has no solutions.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−2	0	−1	0	−8
0	1	0	0	−4	7
0	0	0	1	4	5
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=

x2 is free

x3=

x4 is free

x5 is free

x1=

x2=

x3 is free

x4=

x5=

x1=11+4x5

x2=7+4x5

x3 is free

x4=5−4x5

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−5	0	−1	0	−4
0	1	0	0	−3	1
0	0	0	1	3	2
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=3+12x5

x2=1+3x5

x3 is free

x4=2−3x5

x5 is free

x1=

x2=

x3 is free

x4=

x5=

x1=

x2 is free

x3=

x4 is free

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−2	0	−1	0	−9
0	1	0	0	−3	5
0	0	0	1	8	9
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=10−2x5

x2=5+3x5

x3 is free

x4=9−8x5

x5 is free

x1=

x2=

x3 is free

x4=

x5=

x1=

x2 is free

x3=

x4 is free

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−7	0	−1	0	−3
0	1	0	0	−2	3
0	0	0	1	7	2
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=

x2 is free

x3=

x4 is free

x5 is free

x1=

x2=

x3 is free

x4=

x5=

x1=20+7x5

x2=3+2x5

x3 is free

x4=2−7x5

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−7	0	−1	0	−2
0	1	0	0	−3	1
0	0	0	1	4	5
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=

x2=

x3 is free

x4=

x5=

x1=10+17x5

x2=1+3x5

x3 is free

x4=5−4x5

x5 is free

x1=

x2 is free

x3=

x4 is free

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−4	0	−1	0	−6
0	1	0	0	−2	1
0	0	0	1	6	7
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=

x2=

x3 is free

x4=

x5=

x1=5+2x5

x2=1+2x5

x3 is free

x4=7−6x5

x5 is free

x1=

x2 is free

x3=

x4 is free

x5 is free

The system is inconsistent.

1.2.13

Find the general solution of the system whose augmented matrix is given below.

1	−3	0	−1	0	−6
0	1	0	0	−9	7
0	0	0	1	8	5
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your answer.

x1=

x2=

x3 is free

x4=

x5=

x1=20+19x5

x2=7+9x5

x3 is free

x4=5−8x5

x5 is free

x1=

x2 is free

x3=

x4 is free

x5 is free

The system is inconsistent.

1.2.14

Find the general solution of the system whose augmented matrix is given below.

1	0	−5	0	−6	4
0	1	8	−1	0	3
0	0	0	0	1	0
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=

x2=

x3=

x4=

x5 is free

x1=

x2=

x3 is free

x4=

x5=

x1=4+5x3

x2=3+x4−8x3

x3 is free

x4 is free

x5=0

The system is inconsistent.

1.2.14

Find the general solution of the system whose augmented matrix is given below.

1	0	−5	0	−4	8
0	1	6	−1	0	7
0	0	0	0	1	0
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=8+5x3

x2=7+x4−6x3

x3 is free

x4 is free

x5=0

x1=

x2=

x3=

x4=

x5 is free

x1=

x2=

x3 is free

x4=

x5=

The system is inconsistent.

1.2.14

Find the general solution of the system whose augmented matrix is given below.

1	0	−5	0	−4	7
0	1	3	−1	0	2
0	0	0	0	1	0
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=

x2=

x3 is free

x4=

x5=

x1=7+5x3

x2=2+x4−3x3

x3 is free

x4 is free

x5=0

x1=

x2=

x3=

x4=

x5 is free

The system is inconsistent.

1.2.14

Find the general solution of the system whose augmented matrix is given below.

1	0	−6	0	−7	4
0	1	3	−1	0	5
0	0	0	0	1	0
0	0	0	0	0	0

Select the correct choice below and, if necessary, fill in any answer boxes to complete your answer.

x1=4+6x3

x2=5+x4−3x3

x3 is free

x4 is free

x5=0

x1=

x2=

x3=

x4=

x5 is free

x1=

x2=

x3 is free

x4=

x5=

The system is inconsistent.

1.2.22

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

1	h	2
−4	20	−10

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrix is the augmented matrix of a consistent linear system if

h=.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system if

h≠−5.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system for every value of h.

The matrix is not the augmented matrix of a consistent linear system for any value of h.

1.2.22

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

1	h	3
2	10	3

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrix is the augmented matrix of a consistent linear system if

h≠5.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system if

h=.

(Use a comma to separate answers as needed. Type an integer or a simplified fraction.)

The matrix is the augmented matrix of a consistent linear system for every value of h.

The matrix is not the augmented matrix of a consistent linear system for any value of h.

1.2.25

Determine whether the statement below is true or false. Justify the answer.

In some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations.

Choose the correct answer below.

The statement is false. For each matrix, there is only one sequence of row operations that row reduces it.

The statement is true. The echelon form of a matrix is always unique, but the reduced echelon form of a matrix might not be unique.

The statement is true. It is possible for there to be several different sequences of row operations that row reduce a matrix.

The statement is false. Each matrix is row equivalent to one and only one reduced echelon matrix.

1.2.25

Determine whether the statement below is true or false. Justify the answer.

In some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations.

Choose the correct answer below.

The statement is false. Each matrix is row equivalent to one and only one reduced echelon matrix.

The statement is true. It is possible for there to be several different sequences of row operations that row reduce a matrix.

The statement is false. For each matrix, there is only one sequence of row operations that row reduces it.

The statement is true. The echelon form of a matrix is always unique, but the reduced echelon form of a matrix might not be unique.

1.2.26

Determine whether the statement below is true or false. Justify the answer.

The echelon form of a matrix is unique.

Choose the correct answer below.

The statement is false. The echelon form of a matrix is not unique, but the reduced echelon form is unique.

The statement is true. Both the echelon form and the reduced echelon form of a matrix are unique. They are the same regardless of the chosen row operations.

The statement is false. Neither the echelon form nor the reduced echelon form of a matrix are unique. They depend on the row operations performed.

The statement is true. The echelon form of a matrix is always unique, but the reduced echelon form of a matrix might not be unique.

1.2.26

Determine whether the statement below is true or false. Justify the answer.

The echelon form of a matrix is unique.

Choose the correct answer below.

The statement is true. The echelon form of a matrix is always unique, but the reduced echelon form of a matrix might not be unique.

The statement is false. Neither the echelon form nor the reduced echelon form of a matrix are unique. They depend on the row operations performed.

The statement is false. The echelon form of a matrix is not unique, but the reduced echelon form is unique.

The statement is true. Both the echelon form and the reduced echelon form of a matrix are unique. They are the same regardless of the chosen row operations.

1.2.26

Determine whether the statement below is true or false. Justify the answer.

The echelon form of a matrix is unique.

Choose the correct answer below.

The statement is true. The echelon form of a matrix is always unique, but the reduced echelon form of a matrix might not be unique.

The statement is true. Both the echelon form and the reduced echelon form of a matrix are unique. They are the same regardless of the chosen row operations.

The statement is false. The echelon form of a matrix is not unique, but the reduced echelon form is unique.

The statement is false. Neither the echelon form nor the reduced echelon form of a matrix are unique. They depend on the row operations performed.

1.2.29

Determine whether the statement below is true or false. Justify the answer.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.

Choose the correct answer below.

The statement is true. If a linear system has both basic and free variables, then each basic variable can be expressed in terms of the free variables.

The statement is true. It is the definition of a basic variable.

The statement is false. A variable that corresponds to a pivot column in the coefficient matrix is called a free variable, not a basic variable.

The statement is false. Not every linear system has basic variables.

1.2.29

Determine whether the statement below is true or false. Justify the answer.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.

Choose the correct answer below.

The statement is true. If a linear system has both basic and free variables, then each basic variable can be expressed in terms of the free variables.

The statement is false. Not every linear system has basic variables.

The statement is true. It is the definition of a basic variable.

The statement is false. A variable that corresponds to a pivot column in the coefficient matrix is called a free variable, not a basic variable.

1.2.29

Determine whether the statement below is true or false. Justify the answer.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.

Choose the correct answer below.

The statement is false. Not every linear system has basic variables.

The statement is true. It is the definition of a basic variable.

The statement is true. If a linear system has both basic and free variables, then each basic variable can be expressed in terms of the free variables.

The statement is false. A variable that corresponds to a pivot column in the coefficient matrix is called a free variable, not a basic variable.

1.2.29

Determine whether the statement below is true or false. Justify the answer.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.

Choose the correct answer below.

The statement is false. Not every linear system has basic variables.

The statement is true. It is the definition of a basic variable.

The statement is false. A variable that corresponds to a pivot column in the coefficient matrix is called a free variable, not a basic variable.

The statement is true. If a linear system has both basic and free variables, then each basic variable can be expressed in terms of the free variables.

1.2.31

Determine whether the statement below is true or false. Justify the answer.

Finding a parametric description of the solution set of a linear system is the same as solving the system.

Choose the correct answer below.

The statement is true. Solving a linear system is the same as finding the solution set of the system. The solution set of a linear system can always be expressed using a parametric description.

The statement is false. The solution set of a linear system can only be expressed using a parametric description if the system has no more than one solution.

The statement is true. Regardless of whether a linear system has free variables, the solution set of the system can be expressed using a parametric description.

The statement is false. The solution set of a linear system can only be expressed using a parametric description if the system has at least one solution.

1.2.31

Determine whether the statement below is true or false. Justify the answer.

Finding a parametric description of the solution set of a linear system is the same as solving the system.

Choose the correct answer below.

The statement is false. The solution set of a linear system can only be expressed using a parametric description if the system has no more than one solution.

The statement is true. Solving a linear system is the same as finding the solution set of the system. The solution set of a linear system can always be expressed using a parametric description.

The statement is false. The solution set of a linear system can only be expressed using a parametric description if the system has at least one solution.

The statement is true. Regardless of whether a linear system has free variables, the solution set of the system can be expressed using a parametric description.

1.2.34

Determine whether the statement below is true or false. Justify the answer.

A general solution of a system is an explicit description of all solutions of the system.

Choose the correct answer below.

The statement is true. After applying the row reduction algorithm and generating a general solution of a system, the rightmost column displays all of the particular solutions of that system.

The statement is true. The row reduction algorithm leads directly to an explicit description of the solution set of a linear system when the algorithm is applied to the augmented matrix of the system, leading to a general solution of a system.

The statement is false. A general solution is the result of an inconsistent system, which has no particular solution.

The statement is false. Each different choice of a free variable produces the same solution of the system.

1.2.34

Determine whether the statement below is true or false. Justify the answer.

A general solution of a system is an explicit description of all solutions of the system.

Choose the correct answer below.

The statement is true. After applying the row reduction algorithm and generating a general solution of a system, the rightmost column displays all of the particular solutions of that system.

The statement is true. The row reduction algorithm leads directly to an explicit description of the solution set of a linear system when the algorithm is applied to the augmented matrix of the system, leading to a general solution of a system.

The statement is false. A general solution is the result of an inconsistent system, which has no particular solution.

The statement is false. Each different choice of a free variable produces the same solution of the system.

1.2.35

Suppose a

4×6

coefficient matrix for a system has

four

pivot columns. Is the system consistent? Why or why not?

Choose the correct answer below.

There is a pivot position in each row of the coefficient matrix. The augmented matrix will have

seven

columns and will not have a row of the form

0

0

0

0

0

0

1

,

so the system is consistent.

There is at least one row of the coefficient matrix that does not have a pivot position. This means the augmented matrix, which will have

seven

columns, must have a row of the form

0

0

0

0

0

0

1

,

so the system is inconsistent.

There is at least one row of the coefficient matrix that does not have a pivot position. This means the augmented matrix, which will have

seven

columns, could have a row of the form

0

0

0

0

0

0

1

,

so the system could be inconsistent.

There is a pivot position in each row of the coefficient matrix. The augmented matrix will have

five

columns and will not have a row of the form

0

0

0

0

1

,

so the system is consistent.

1.2.35

Suppose a

5×8

coefficient matrix for a system has

five

pivot columns. Is the system consistent? Why or why not?

Choose the correct answer below.

There is a pivot position in each row of the coefficient matrix. The augmented matrix will have

six

columns and will not have a row of the form

0

0

0

0

0

1

,

so the system is consistent.

There is at least one row of the coefficient matrix that does not have a pivot position. This means the augmented matrix, which will have

nine

columns, could have a row of the form

0

0

0

0

0

0

0

0

1

,

so the system could be inconsistent.

There is a pivot position in each row of the coefficient matrix. The augmented matrix will have

nine

columns and will not have a row of the form

0

0

0

0

0

0

0

0

1

,

so the system is consistent.

There is at least one row of the coefficient matrix that does not have a pivot position. This means the augmented matrix, which will have

nine

columns, must have a row of the form

0

0

0

0

0

0

0

0

1

,

so the system is inconsistent.

1.2.36

Suppose a system of linear equations has a

3×5

augmented matrix whose fifth column is not a pivot column. Is the system consistent? Why or why not?

To determine if the linear system is consistent, use the portion of the Existence and Uniqueness Theorem, shown below.

A linear system is consistent if and only if the rightmost column of the augmented matrix

is not

a pivot column. That is, if and only if an echelon form of the augmented matrix has

no row

of the form

	0	⋯	0	b

with b nonzero.

In the augmented matrix described above, is the rightmost column a pivot column?

No

Yes

In the echelon form of the augmented matrix, is there a row of the form

	0	0	0	0	b

with b nonzero?

No

Yes

Therefore, by the Existence and Uniqueness Theorem, the linear system is

consistent.

1.2.36

Suppose a system of linear equations has a

3×5

augmented matrix whose fifth column is not a pivot column. Is the system consistent? Why or why not?

To determine if the linear system is consistent, use the portion of the Existence and Uniqueness Theorem, shown below.

A linear system is consistent if and only if the rightmost column of the augmented matrix

is not

a pivot column. That is, if and only if an echelon form of the augmented matrix has

no row

of the form

	0	⋯	0	b

with b nonzero.

In the augmented matrix described above, is the rightmost column a pivot column?

Yes

No

In the echelon form of the augmented matrix, is there a row of the form

	0	0	0	0	b

with b nonzero?

Yes

No

Therefore, by the Existence and Uniqueness Theorem, the linear system is

consistent.

1.2.37

Suppose the coefficient matrix of a system of linear equations has a pivot position in every row. Explain why the system is consistent.

Choose the correct answer below.

The system is consistent because the augmented matrix will contain a row of the form

0

⋯

0

b

with b nonzero.

The system is consistent because the augmented matrix is row equivalent to one and only one reduced echelon matrix.

The system is consistent because the rightmost column of the augmented matrix is not a pivot column.

The system is consistent because all the columns in the augmented matrix will have a pivot position.

1.2.37

Suppose the coefficient matrix of a system of linear equations has a pivot position in every row. Explain why the system is consistent.

Choose the correct answer below.

The system is consistent because the rightmost column of the augmented matrix is not a pivot column.

The system is consistent because all the columns in the augmented matrix will have a pivot position.

The system is consistent because the augmented matrix will contain a row of the form

0

⋯

0

b

with b nonzero.

The system is consistent because the augmented matrix is row equivalent to one and only one reduced echelon matrix.

1.2.45

Suppose experimental data are represented by a set of points in the plane. An interpolating polynomial for the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. One method for finding an interpolating polynomial is to solve a system of linear equations. Find the interpolating polynomial

p(t)=a0+a1t+a2t2

for the data

(1,12),

(2,15),

(3,16).

That is, find

a0,

a1,

and

a2

such that the following is true.

a0	+	a1(1)	+	a2(1)2	=	12
a0	+	a1(2)	+	a2(2)2	=	15
a0	+	a1(3)	+	a2(3)2	=	16

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The interpolating polynomial is

p(t)=7+6t−t2.

There are infinitely many possible interpolating polynomials.

There does not exist an interpolating polynomial for the given data.

1.2.45

Suppose experimental data are represented by a set of points in the plane. An interpolating polynomial for the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. One method for finding an interpolating polynomial is to solve a system of linear equations. Find the interpolating polynomial

p(t)=a0+a1t+a2t2

for the data

(1,11),

(2,19),

(3,21).

That is, find

a0,

a1,

and

a2

such that the following is true.

a0	+	a1(1)	+	a2(1)2	=	11
a0	+	a1(2)	+	a2(2)2	=	19
a0	+	a1(3)	+	a2(3)2	=	21

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The interpolating polynomial is

p(t)=−3+17t−3t2.

There are infinitely many possible interpolating polynomials.

There does not exist an interpolating polynomial for the given data.

1.2.45

Suppose experimental data are represented by a set of points in the plane. An interpolating polynomial for the data is a polynomial whose graph passes through every point. In scientific work, such a polynomial can be used, for example, to estimate values between the known data points. Another use is to create curves for graphical images on a computer screen. One method for finding an interpolating polynomial is to solve a system of linear equations. Find the interpolating polynomial

p(t)=a0+a1t+a2t2

for the data

(1,14),

(2,18),

(3,20).

That is, find

a0,

a1,

and

a2

such that the following is true.

a0	+	a1(1)	+	a2(1)2	=	14
a0	+	a1(2)	+	a2(2)2	=	18
a0	+	a1(3)	+	a2(3)2	=	20

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The interpolating polynomial is

p(t)=8+7t−t2.

There are infinitely many possible interpolating polynomials.

There does not exist an interpolating polynomial for the given data.

1.3.5

Write a system of equations that is equivalent to the given vector equation.

x1

	2
	−4
	6

+x2

	6
	0
	−8

=

	4
	−5
	7

Choose the correct answer below.

2x1	+	6x2	=	4
−4x1	+	x2	=	−5
6x1	−	8x2	=	7

2x1	+	6x2	=	7
−4x1			=	−5
6x1	−	8x2	=	4

2x1	+	6x2	=	4
−4x1			=	−5
6x1	−	8x2	=	7

2x1	+	6x2	=	−5
−4x1			=	4
6x1	−	8x2	=	7

1.3.5

Write a system of equations that is equivalent to the given vector equation.

x1

	4
	−4
	7

+x2

	6
	0
	−8

=

	2
	−2
	9

Choose the correct answer below.

4x1	+	6x2	=	2
−4x1	+	x2	=	−2
7x1	−	8x2	=	9

4x1	+	6x2	=	2
−4x1			=	−2
7x1	−	8x2	=	9

4x1	+	6x2	=	−2
−4x1			=	2
7x1	−	8x2	=	9

4x1	+	6x2	=	9
−4x1			=	−2
7x1	−	8x2	=	2

1.3.6

Write a system of equations that is equivalent to the given vector equation.

x1

5

−2

+x2

6

4

+x3

−5

1

=

0

0

Choose the correct answer below.

5x1	+	6x2	+	5x3	=	0
2x1	+	4x2	+	x3	=	0

5x1	+	6x2	−	5x3	=	0
−2x1	+	4x2			=	0

5x1	+	6x2	+	5x3	=	0
2x1	+	4x2			=	0

5x1	+	6x2	−	5x3	=	0
−2x1	+	4x2	+	x3	=	0

1.3.6

Write a system of equations that is equivalent to the given vector equation.

x1

4

−3

+x2

7

4

+x3

−2

1

=

0

0

Choose the correct answer below.

4x1	+	7x2	−	2x3	=	0
−3x1	+	4x2	+	x3	=	0

4x1	+	7x2	+	2x3	=	0
3x1	+	4x2	+	x3	=	0

4x1	+	7x2	+	2x3	=	0
3x1	+	4x2			=	0

4x1	+	7x2	−	2x3	=	0
−3x1	+	4x2			=	0

1.3.9

Write a vector equation that is equivalent to the given system of equations.

		x2	+	3x3	=	0
5x1	+	9x2	−	x3	=	0
−x1	+	4x2	−	6x3	=	0

Choose the correct answer below.

x1

0

5

−1

+x2

1

9

4

+x3

0

0

0

=

3

−1

−6

x1

1

9

4

+x2

0

5

−1

+x3

3

−1

−6

=

0

0

0

x1

0

5

−1

+x2

3

−1

−6

+x3

1

9

4

=

0

0

0

x1

0

5

−1

+x2

1

9

4

+x3

3

−1

−6

=

0

0

0

1.3.9

Write a vector equation that is equivalent to the given system of equations.

		x2	+	2x3	=	0
4x1	+	6x2	−	x3	=	0
−x1	+	3x2	−	8x3	=	0

Choose the correct answer below.

x1

0

4

−1

+x2

2

−1

−8

+x3

1

6

3

=

0

0

0

x1

0

4

−1

+x2

1

6

3

+x3

2

−1

−8

=

0

0

0

x1

0

4

−1

+x2

1

6

3

+x3

0

0

0

=

2

−1

−8

x1

1

6

3

+x2

0

4

−1

+x3

2

−1

−8

=

0

0

0

1.3.9

Write a vector equation that is equivalent to the given system of equations.

		x2	+	2x3	=	0
5x1	+	8x2	−	x3	=	0
−x1	+	3x2	−	8x3	=	0

Choose the correct answer below.

x1

1

8

3

+x2

0

5

−1

+x3

2

−1

−8

=

0

0

0

x1

0

5

−1

+x2

1

8

3

+x3

0

0

0

=

2

−1

−8

x1

0

5

−1

+x2

2

−1

−8

+x3

1

8

3

=

0

0

0

x1

0

5

−1

+x2

1

8

3

+x3

2

−1

−8

=

0

0

0

1.3.9

Write a vector equation that is equivalent to the given system of equations.

		x2	+	5x3	=	0
5x1	+	6x2	−	x3	=	0
−x1	+	3x2	−	8x3	=	0

Choose the correct answer below.

x1

0

5

−1

+x2

5

−1

−8

+x3

1

6

3

=

0

0

0

x1

0

5

−1

+x2

1

6

3

+x3

5

−1

−8

=

0

0

0

x1

0

5

−1

+x2

1

6

3

+x3

0

0

0

=

5

−1

−8

x1

1

6

3

+x2

0

5

−1

+x3

5

−1

−8

=

0

0

0

1.3.11

Determine if b is a linear combination of

a1,

a2,

and

a3.

a1=

1

−3

0

,

a2=

0

1

4

,

a3=

5

−6

36

,

b=

2

−1

20

Choose the correct answer below.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column.

Vector b is not a linear combination of

a1,

a2,

and

a3.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column and the second entry in the second column.

1.3.11

Determine if b is a linear combination of

a1,

a2,

and

a3.

a1=

1

−4

0

,

a2=

0

1

3

,

a3=

5

−6

42

,

b=

3

−2

30

Choose the correct answer below.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column and the second entry in the second column.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column.

Vector b is not a linear combination of

a1,

a2,

and

a3.

1.3.11

Determine if b is a linear combination of

a1,

a2,

and

a3.

a1=

1

−3

0

,

a2=

0

1

2

,

a3=

5

−5

20

,

b=

4

−1

22

Choose the correct answer below.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column.

Vector b is not a linear combination of

a1,

a2,

and

a3.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column and the second entry in the second column.

1.3.11

Determine if b is a linear combination of

a1,

a2,

and

a3.

a1=

1

−2

0

,

a2=

0

1

4

,

a3=

6

−5

28

,

b=

3

−2

16

Choose the correct answer below.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column and the second entry in the second column.

Vector b is not a linear combination of

a1,

a2,

and

a3.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.

Vector b is a linear combination of

a1,

a2,

and

a3.

The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column.

1.3.13

Determine if b is a linear combination of the vectors formed from the columns of the matrix A.

A=

1	−3	4
0	2	5
−3	9	−12

,

b=

4

−5

−4

Choose the correct answer below.

Vector b is a linear combination of the vectors formed from the columns of the matrix A. The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column.

Vector b is a linear combination of the vectors formed from the columns of the matrix A. The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column.

Vector b is not a linear combination of the vectors formed from the columns of the matrix A.

Vector b is a linear combination of the vectors formed from the columns of the matrix A. The pivots in the corresponding echelon matrix are in the first entry in the first column and the third entry in the second column, and the third entry in the third column.

1.3.15

List five vectors in

Spanv1, v2.

Do not make a sketch.

v1=

	6
	1
	−7

,

v2=

	−5
	3
	0

List five vectors in

Spanv1, v2.

	0
	0
	0

,

	6
	1
	−7

,

	−5
	3
	0

,

	1
	4
	−7

,

	11
	−2
	−7

(Use the matrix template in the math palette. Use a comma to separate vectors as needed. Type an integer or a simplified fraction for each vector element. Type each answer only once.)

1.3.15

List five vectors in

Spanv1, v2.

Do not make a sketch.

v1=

	6
	2
	−7

,

v2=

	−4
	4
	0

List five vectors in

Spanv1, v2.

	0
	0
	0

,

	6
	2
	−7

,

	−4
	4
	0

,

	2
	6
	−7

,

	10
	−2
	−7

(Use the matrix template in the math palette. Use a comma to separate vectors as needed. Type an integer or a simplified fraction for each vector element. Type each answer only once.)

1.3.27

Determine whether the statement below is true or false. Justify the answer.

An example of a linear combination of vectors

v1

and

v2

is the vector

12v1.

Choose the correct answer below.

The statement is true because

12v1=14v1+14v2.

The statement is false because

12v1

cannot be expressed as a multiple of

v2.

The statement is true because

12v1=12v1+0v2.

The statement is false because a linear combination of

v1

and

v2

needs to involve both of those vectors.

1.3.32

Determine whether the statement below is true or false. Justify the answer.

Asking whether the linear system corresponding to an augmented matrix

	a1	a2	a3	b

has a solution amounts to asking whether

b

is in

Span a1, a2, a3.

Choose the correct answer below.

The statement is false. If

b

corresponds to the origin, then it cannot be in

Span a1, a2, a3.

The statement is true. The solution of the linear system corresponding to

	a1	a2	a3	b

is always in

Span a1, a2, a3.

The statement is true. The linear system corresponding to

	a1	a2	a3	b

has a solution when

b

can be written as a linear combination of

a1,

a2,

and

a3.

This is equivalent to saying that

b

is in

Span a1, a2, a3.

The statement is false. It is possible for the linear system corresponding to

	a1	a2	a3	b

to have a solution without

b

being in

Span a1, a2, a3.

1.3.32

Determine whether the statement below is true or false. Justify the answer.

Asking whether the linear system corresponding to an augmented matrix

	a1	a2	a3	b

has a solution amounts to asking whether

b

is in

Span a1, a2, a3.

Choose the correct answer below.

The statement is true. The linear system corresponding to

	a1	a2	a3	b

has a solution when

b

can be written as a linear combination of

a1,

a2,

and

a3.

This is equivalent to saying that

b

is in

Span a1, a2, a3.

The statement is false. If

b

corresponds to the origin, then it cannot be in

Span a1, a2, a3.

The statement is true. The solution of the linear system corresponding to

	a1	a2	a3	b

is always in

Span a1, a2, a3.

The statement is false. It is possible for the linear system corresponding to

	a1	a2	a3	b

to have a solution without

b

being in

Span a1, a2, a3.

1.3.32

Determine whether the statement below is true or false. Justify the answer.

Asking whether the linear system corresponding to an augmented matrix

	a1	a2	a3	b

has a solution amounts to asking whether

b

is in

Span a1, a2, a3.

Choose the correct answer below.

The statement is false. If

b

corresponds to the origin, then it cannot be in

Span a1, a2, a3.

The statement is true. The solution of the linear system corresponding to

	a1	a2	a3	b

is always in

Span a1, a2, a3.

The statement is true. The linear system corresponding to

	a1	a2	a3	b

has a solution when

b

can be written as a linear combination of

a1,

a2,

and

a3.

This is equivalent to saying that

b

is in

Span a1, a2, a3.

The statement is false. It is possible for the linear system corresponding to

	a1	a2	a3	b

to have a solution without

b

being in

Span a1, a2, a3.

1.3.32

Determine whether the statement below is true or false. Justify the answer.

Asking whether the linear system corresponding to an augmented matrix

	a1	a2	a3	b

has a solution amounts to asking whether

b

is in

Span a1, a2, a3.

Choose the correct answer below.

The statement is false. It is possible for the linear system corresponding to

	a1	a2	a3	b

to have a solution without

b

being in

Span a1, a2, a3.

The statement is true. The linear system corresponding to

	a1	a2	a3	b

has a solution when

b

can be written as a linear combination of

a1,

a2,

and

a3.

This is equivalent to saying that

b

is in

Span a1, a2, a3.

The statement is true. The solution of the linear system corresponding to

	a1	a2	a3	b

is always in

Span a1, a2, a3.

The statement is false. If

b

corresponds to the origin, then it cannot be in

Span a1, a2, a3.

1.3.33

Let

A=

	1	0	−6
	0	3	−5
	−6	9	5

and

b=

	4
	−4
	−20

.

Denote the columns of A by

a1,

a2,

a3,

and let

W=Span a1, a2, a3.

a. Is

b

in

a1, a2, a3?

How many vectors are in

a1, a2, a3?

b. Is

b

in W? How many vectors are in W?

c. Show that

a1

is in W.

[Hint:

Row operations are unnecessary.]

a. Is

b

in

a1, a2, a3?

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

Yes,

b

is in

a1, a2, a3

since, although

b

is not equal to

a1,

a2,

or

a3,

it can be expressed as a linear combination of them. In particular,

b=a1+a2+a3.

(Simplify your answers.)

Yes,

b

is in

a1, a2, a3

since

b=a.

(Type a whole number.)

No,

b

is not in

a1, a2, a3

since it cannot be generated by a linear combination of

a1,

a2,

and

a3.

No,

b

is not in

a1, a2, a3

since

b

is not equal to

a1,

a2,

or

a3.

How many vectors are in

a1, a2, a3?

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

There is(are)

3

vector(s) in

a1, a2, a3.

(Type a whole number.)

There are infinitely many vectors in

a1, a2, a3.

b. Is

b

in W? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

No,

b

is not in W since

b

is not equal to

a1,

a2,

or

a3.

Yes,

b

is in W since

b=a.

(Type a whole number.)

Yes,

b

is in W since, although

b

is not equal to

a1,

a2,

or

a3,

it can be expressed as a linear combination of them. In particular,

b=−2a1+−3a2+−1a3.

(Simplify your answers.)

No,

b

is not in W since it cannot be generated by a linear combination of

a1,

a2,

and

a3.

How many vectors are in W? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

There is(are)

vector(s) in W.

(Type a whole number.)

There are infinitely many vectors in W.

c. The vector

a1

is in W if and only if

a1

can be written in the form

c1a1+c2a2+c3a3

with

c1, c2, and c3 all being scalars.

The vector

a1

can be written as the linear combination

1a1+0a2+0a3.

This satisfies the condition identified in the previous step, so it follows that

a1

is in W.

(Simplify your answers.)

1.3.33

Let

A=

	1	0	−5
	0	2	−3
	−4	4	5

and

b=

	7
	4
	−2

.

Denote the columns of A by

a1,

a2,

a3,

and let

W=Span a1, a2, a3.

a. Is

b

in

a1, a2, a3?

How many vectors are in

a1, a2, a3?

b. Is

b

in W? How many vectors are in W?

c. Show that

a2

is in W.

[Hint:

Row operations are unnecessary.]

a. Is

b

in

a1, a2, a3?

Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

No,

b

is not in

a1, a2, a3

since it cannot be generated by a linear combination of

a1,

a2,

and

a3.

Yes,

b

is in

a1, a2, a3

since

b=a.

(Type a whole number.)

Yes,

b

is in

a1, a2, a3

since, although

b

is not equal to

a1,

a2,

or

a3,

it can be expressed as a linear combination of them. In particular,

b=a1+a2+a3.

(Simplify your answers.)

No,

b

is not in

a1, a2, a3

since

b

is not equal to

a1,

a2,

or

a3.

How many vectors are in

a1, a2, a3?

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

There is(are)

3

vector(s) in

a1, a2, a3.

(Type a whole number.)

There are infinitely many vectors in

a1, a2, a3.

b. Is

b

in W? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

No,

b

is not in W since it cannot be generated by a linear combination of

a1,

a2,

and

a3.

No,

b

is not in W since

b

is not equal to

a1,

a2,

or

a3.

Yes,

b

is in W since

b=a.

(Type a whole number.)

Yes,

b

is in W since, although

b

is not equal to

a1,

a2,

or

a3,

it can be expressed as a linear combination of them. In particular,

b=−3a1+−1a2+−2a3.

(Simplify your answers.)

How many vectors are in W? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

There is(are)

vector(s) in W.

(Type a whole number.)

There are infinitely many vectors in W.

c. The vector

a2

is in W if and only if

a2

can be written in the form

c1a1+c2a2+c3a3

with

c1, c2, and c3 all being scalars.

The vector

a2

can be written as the linear combination

0a1+1a2+0a3.

This satisfies the condition identified in the previous step, so it follows that

a2

is in W.

(Simplify your answers.)

1.4.7

Use the definition of

Ax

to write the vector equation as a matrix equation.

x1

−9

8

4

−6

+x2

−1

8

−9

4

+x3

6

−8

−7

−4

=

5

3

1

7

The matrix equation is

	−9	−1	6
	8	8	−8
	4	−9	−7
	−6	4	−4

	x1
	x2
	x3

=

	5
	3
	1
	7

1.4.7

Use the definition of

Ax

to write the vector equation as a matrix equation.

x1

2

5

−6

−2

+x2

0

−4

−4

−7

+x3

−2

7

2

−2

=

9

8

1

3

The matrix equation is

	2	0	−2
	5	−4	7
	−6	−4	2
	−2	−7	−2

	x1
	x2
	x3

=

	9
	8
	1
	3

1.4.8

Use the definition of

Ax

to write the vector equation as a matrix equation.

z1

	−1
	1

+z2

	−5
	−5

+z3

	−4
	0

+z4

	5
	5

=

	3
	−1

The vector equation written as a matrix equation is

	−1	−5	−4	5
	1	−5	0	5

	z1
	z2
	z3
	z4

=

	3
	−1

.

1.4.8

Use the definition of

Ax

to write the vector equation as a matrix equation.

z1

	−4
	3

+z2

	−2
	−1

+z3

	−3
	5

+z4

	2
	5

=

	3
	18

The vector equation written as a matrix equation is

	−4	−2	−3	2
	3	−1	5	5

	z1
	z2
	z3
	z4

=

	3
	18

.

1.4.9

Write the system first as a vector equation and then as a matrix equation.

8x1	+	x2	−	3x3	=	8
		6x2	+	10x3	=	0

Write the system as a vector equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	x1
	x2
	x3

=

x1

	8
	0

+x2

	1
	6

+x3

	−3
	10

=

	8
	0

	x1	x2	x3

=

Write the system as a matrix equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	x1	x2	x3

=

	8	1	−3
	0	6	10

	x1
	x2
	x3

=

	8
	0

x1+x2+x3=

1.4.10

Write the system first as a vector equation and then as a matrix equation.

9x1	−	x2	=	5
4x1	+	5x2	=	4
10x1	−	x2	=	1

Write the system as a vector equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	x1
	x2

=

	x1	x2

=

x1

	9
	4
	10

+x2

	−1
	5
	−1

=

	5
	4
	1

Write the system as a matrix equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	x1	x2

=

	9	−1
	4	5
	10	−1

	x1
	x2

=

	5
	4
	1

x1+x2=

1.4.10

Write the system first as a vector equation and then as a matrix equation.

2x1	−	x2	=	10
5x1	+	6x2	=	4
9x1	−	x2	=	1

Write the system as a vector equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	x1
	x2

=

x1

	2
	5
	9

+x2

	−1
	6
	−1

=

	10
	4
	1

	x1	x2

=

Write the system as a matrix equation where the first equation of the system corresponds to the first row. Select the correct choice below and fill in any answer boxes to complete your choice.

	2	−1
	5	6
	9	−1

	x1
	x2

=

	10
	4
	1

x1+x2=

	x1	x2

=

1.4.11

Given A and b to the right, write the augmented matrix for the linear system that corresponds to the matrix equation Ax=b. Then solve the system and write the solution as a vector.

A= 1 2 −3 −2 −5 2 −3 −2 4 , b= 4 −13 8

Write the augmented matrix for the linear system that corresponds to the matrix equation

Ax=b.

	1	2	−3	4
	−2	−5	2	−13
	−3	−2	4	8

Solve the system and write the solution as a vector.

x=

	x1
	x2
	x3

=

	−6
	5
	0

1.4.11

Given A and b to the right, write the augmented matrix for the linear system that corresponds to the matrix equation Ax=b. Then solve the system and write the solution as a vector.

A= 1 4 −3 −2 −2 0 3 2 4 , b= 2 14 −24

Write the augmented matrix for the linear system that corresponds to the matrix equation

Ax=b.

	1	4	−3	2
	−2	−2	0	14
	3	2	4	−24

Solve the system and write the solution as a vector.

x=

	x1
	x2
	x3

=

	−10
	3
	0

1.4.22

Let

v1=

0

0

−4

,

v2=

0

−3

−8

,

and

v3=

5

−2

−12

.

Does

v1,v2,v3

span

ℝ3?

Why or why not?

Choose the correct answer below.

Yes. Any vector in

ℝ3

except the zero vector can be written as a linear combination of these three vectors.

No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only two rows.

Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.

No. The set of given vectors spans a plane in

ℝ3.

Any of the three vectors can be written as a linear combination of the other two.

1.4.22

Let

v1=

0

0

−2

,

v2=

0

−6

−6

,

and

v3=

6

−4

−4

.

Does

v1,v2,v3

span

ℝ3?

Why or why not?

Choose the correct answer below.

No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only two rows.

No. The set of given vectors spans a plane in

ℝ3.

Any of the three vectors can be written as a linear combination of the other two.

Yes. Any vector in

ℝ3

except the zero vector can be written as a linear combination of these three vectors.

Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.

1.4.22

Let

v1=

0

0

−4

,

v2=

0

−4

−8

,

and

v3=

4

−3

12

.

Does

v1,v2,v3

span

ℝ3?

Why or why not?

Choose the correct answer below.

No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only two rows.

No. The set of given vectors spans a plane in

ℝ3.

Any of the three vectors can be written as a linear combination of the other two.

Yes. Any vector in

ℝ3

except the zero vector can be written as a linear combination of these three vectors.

Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.

1.4.22

Let

v1=

0

0

−2

,

v2=

0

−3

6

,

and

v3=

6

−2

−4

.

Does

v1,v2,v3

span

ℝ3?

Why or why not?

Choose the correct answer below.

No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only two rows.

Yes. Any vector in

ℝ3

except the zero vector can be written as a linear combination of these three vectors.

Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.

No. The set of given vectors spans a plane in

ℝ3.

Any of the three vectors can be written as a linear combination of the other two.

1.4.22

Let

v1=

0

0

−5

,

v2=

0

−2

−10

,

and

v3=

3

−5

−15

.

Does

v1,v2,v3

span

ℝ3?

Why or why not?

Choose the correct answer below.

No. The set of given vectors spans a plane in

ℝ3.

Any of the three vectors can be written as a linear combination of the other two.

Yes. Any vector in

ℝ3

except the zero vector can be written as a linear combination of these three vectors.

No. When the given vectors are written as the columns of a matrix A, A has a pivot position in only two rows.

Yes. When the given vectors are written as the columns of a matrix A, A has a pivot position in every row.

1.4.26

Determine whether the statement below is true or false. Justify the answer.

A vector b is a linear combination of the columns of a matrix A if and only if the equation

Ax=b

has at least one solution.

Choose the correct answer below.

This statement is true. The equation

Ax=b

has the same solution set as the equation

x1a1+x2a2+⋯+xnan=b.

This statement is false. If the matrix A is the identity matrix, then the equation

Ax=b

has at least one solution, but b is not a linear combination of the columns of A.

This statement is false. If the equation

Ax=b

has infinitely many solutions, then the vector b cannot be a linear combination of the columns of A.

This statement is true. The equation

Ax=b

is unrelated to whether the vector b is a linear combination of the columns of a matrix A.

1.4.26

Determine whether the statement below is true or false. Justify the answer.

A vector b is a linear combination of the columns of a matrix A if and only if the equation

Ax=b

has at least one solution.

Choose the correct answer below.

This statement is true. The equation

Ax=b

has the same solution set as the equation

x1a1+x2a2+⋯+xnan=b.

This statement is false. If the equation

Ax=b

has infinitely many solutions, then the vector b cannot be a linear combination of the columns of A.

This statement is true. The equation

Ax=b

is unrelated to whether the vector b is a linear combination of the columns of a matrix A.

This statement is false. If the matrix A is the identity matrix, then the equation

Ax=b

has at least one solution, but b is not a linear combination of the columns of A.

1.4.27

Determine whether the statement below is true or false. Justify the answer.

The equation

Ax=b

is consistent if the augmented matrix

A

b

has a pivot position in every row.

Choose the correct answer below.

This statement is true. The pivot positions in the augmented matrix

A

b

always occur in the columns that represent A.

This statement is false. If the augmented matrix

A

b

has a pivot position in every row, the equation

Ax=b

may or may not be consistent. One pivot position may be in the column representing

b.

This statement is false. The augmented matrix

A

b

cannot have a pivot position in every row because it has more columns than rows.

This statement is true. If the augmented matrix

A

b

has a pivot position in every row, then the equation

Ax=b

has a solution for each b in

ℝm.

1.4.33

Determine whether the statement below is true or false. Justify the answer.

If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then A cannot have a pivot position in every row.

Choose the correct answer below.

This statement is true. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the columns of A span

ℝm.

This statement is false. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the equation

Ax=b

has a solution for each b in

ℝm.

This statement is false. Since the equation

Ax=b

has a solution for each b in

ℝm,

the equation

Ax=b

is consistent for each b in

ℝm.

This statement is true. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the equation

Ax=b

has no solution for some b in

ℝm.

1.4.33

Determine whether the statement below is true or false. Justify the answer.

If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then A cannot have a pivot position in every row.

Choose the correct answer below.

This statement is true. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the equation

Ax=b

has no solution for some b in

ℝm.

This statement is false. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the equation

Ax=b

has a solution for each b in

ℝm.

This statement is true. If A is an

m×n

matrix and if the equation

Ax=b

is inconsistent for some b in

ℝm,

then the columns of A span

ℝm.

This statement is false. Since the equation

Ax=b

has a solution for each b in

ℝm,

the equation

Ax=b

is consistent for each b in

ℝm.

1.4.41

Let A be a

3×2

matrix. Explain why the equation

Ax=b

cannot be consistent for all b in

ℝ3.

Generalize your argument to the case of an arbitrary A with more rows than columns.

Why is the equation

Ax=b

not consistent for all b in

ℝ3?

When written in reduced echelon form, any

3×2

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with a zero on the left side and a possibly nonzero entry of b on the right side.

When written in reduced echelon form, any

3×2

matrix will have at least one column of all zeros. Since there is not a pivot in every column, the matrix cannot be consistent.

When written in reduced echelon form, any

3×2

matrix will have a pivot in every row. Because there are more rows than columns, this is too many pivots, and the system will be inconsistent.

When written in reduced echelon form, any

3×2

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with infinitely many solutions.

Let A be an

m×n

matrix, where

m>n.

Why is

Ax=b

not consistent for all b in

ℝm?

When written in reduced echelon form, any

m×n

matrix will have at least one column of all zeros. Since there is not a pivot in every column, the matrix cannot be consistent.

When written in reduced echelon form, any

m×n

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with a zero on the left side and a possibly nonzero entry of b on the right side.

If A has more rows than columns, the number of pivots cannot be determined without knowing more information about A, and so A cannot be consistent.

When written in reduced echelon form, any

m×n

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with infinitely many solutions.

When written in reduced echelon form, any

m×n

matrix will have a pivot in every row. Because there are more rows than columns, this is too many pivots, and the system will be inconsistent.

1.4.41

Let A be a

3×2

matrix. Explain why the equation

Ax=b

cannot be consistent for all b in

ℝ3.

Generalize your argument to the case of an arbitrary A with more rows than columns.

Why is the equation

Ax=b

not consistent for all b in

ℝ3?

When written in reduced echelon form, any

3×2

matrix will have at least one column of all zeros. Since there is not a pivot in every column, the matrix cannot be consistent.

When written in reduced echelon form, any

3×2

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with a zero on the left side and a possibly nonzero entry of b on the right side.

When written in reduced echelon form, any

3×2

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with infinitely many solutions.

When written in reduced echelon form, any

3×2

matrix will have a pivot in every row. Because there are more rows than columns, this is too many pivots, and the system will be inconsistent.

Let A be an

m×n

matrix, where

m>n.

Why is

Ax=b

not consistent for all b in

ℝm?

When written in reduced echelon form, any

m×n

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with infinitely many solutions.

When written in reduced echelon form, any

m×n

matrix will have a pivot in every row. Because there are more rows than columns, this is too many pivots, and the system will be inconsistent.

When written in reduced echelon form, any

m×n

matrix will have at least one row of all zeros. When solving

Ax=b,

that row will represent an equation with a zero on the left side and a possibly nonzero entry of b on the right side.

When written in reduced echelon form, any

m×n

matrix will have at least one column of all zeros. Since there is not a pivot in every column, the matrix cannot be consistent.

If A has more rows than columns, the number of pivots cannot be determined without knowing more information about A, and so A cannot be consistent.

1.4.44

Suppose A is a

3×3

matrix and b is a vector in

ℝ3

with the property what

Ax=b

has a unique solution. Explain why the columns of A must span

ℝ3.

Choose the correct answer below.

The equation has a unique solution so for each pair of vectors x and b there is only one possible matrix A. Therefore the columns of A must span

ℝ3.

Matrix A is a square matrix, so when computing

Ax,

the row-vector rule shows that the columns of A must span

ℝ3.

When b is written as a linear combination of the columns of A, it simplifies to the vector of weights,

x.

Therefore the columns of A must span

ℝ3.

The reduced echelon form of A must have a pivot in each row or there would be more than one possible solution for the equation

Ax=b.

Therefore the columns of A must span

ℝ3.

1.4.44

Suppose A is a

3×3

matrix and b is a vector in

ℝ3

with the property what

Ax=b

has a unique solution. Explain why the columns of A must span

ℝ3.

Choose the correct answer below.

Matrix A is a square matrix, so when computing

Ax,

the row-vector rule shows that the columns of A must span

ℝ3.

The equation has a unique solution so for each pair of vectors x and b there is only one possible matrix A. Therefore the columns of A must span

ℝ3.

When b is written as a linear combination of the columns of A, it simplifies to the vector of weights,

x.

Therefore the columns of A must span

ℝ3.

The reduced echelon form of A must have a pivot in each row or there would be more than one possible solution for the equation

Ax=b.

Therefore the columns of A must span

ℝ3.

1.5.1

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

2x1−9x2+11x3	=	0
−2x1−11x2+4x3	=	0
4x1+2x2+7x3	=	0

Choose the correct answer below.

It is impossible to determine.

The system has a nontrivial solution.

The system has only a trivial solution.

1.5.1

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

2x1−9x2+11x3	=	0
−2x1−11x2+4x3	=	0
4x1+2x2+7x3	=	0

Choose the correct answer below.

It is impossible to determine.

The system has a nontrivial solution.

The system has only a trivial solution.

1.5.1

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

6x1−3x2+15x3	=	0
−6x1−9x2−6x3	=	0
12x1+6x2+21x3	=	0

Choose the correct answer below.

The system has a nontrivial solution.

The system has only a trivial solution.

It is impossible to determine.

1.5.1

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

2x1−5x2+8x3	=	0
−2x1−7x2+x3	=	0
4x1+2x2+7x3	=	0

Choose the correct answer below.

The system has only a trivial solution.

It is impossible to determine.

The system has a nontrivial solution.

1.5.1

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

4x1−2x2+10x3	=	0
−4x1−2x2−7x3	=	0
8x1+4x2+14x3	=	0

Choose the correct answer below.

The system has a nontrivial solution.

It is impossible to determine.

The system has only a trivial solution.

1.5.2

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

x1	−	3x2	+	6x3	=	0
−3x1	+	4x2	−	10x3	=	0
x1	+	2x2	+	10x3	=	0

Choose the correct answer below.

The system has a nontrivial solution.

The system has only a trivial solution.

It is impossible to determine.

1.5.2

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.

x1	−	2x2	+	6x3	=	0
−4x1	+	3x2	−	15x3	=	0
x1	+	3x2	+	10x3	=	0

Choose the correct answer below.

The system has a nontrivial solution.

The system has only a trivial solution.

It is impossible to determine.

1.5.5

Write the solution set of the given homogeneous system in parametric vector form.

3x1+3x2+6x3	=	0	where the solution set is x= x1 x2 x3
−6x1−6x2−12x3	=	0
−7x2+14x3	=	0

x=x3

	−4
	2
	1

(Type an integer or simplified fraction for each matrix element.)

1.5.5

Write the solution set of the given homogeneous system in parametric vector form.

2x1+2x2+4x3	=	0	where the solution set is x= x1 x2 x3
−4x1−4x2−8x3	=	0
−6x2+18x3	=	0

x=x3

	−5
	3
	1

(Type an integer or simplified fraction for each matrix element.)

1.5.7

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	4	−5	8
	0	1	−5	5

x=x3

	−15
	5
	1
	0

+x4

	12
	−5
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.7

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	3	−5	5
	0	1	−2	8

x=x3

	−1
	2
	1
	0

+x4

	19
	−8
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.8

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	−4	−9	4
	0	1	4	−2

x=x3

	−7
	−4
	1
	0

+x4

	4
	2
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.8

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	−2	−5	3
	0	1	3	−4

x=x3

	−1
	−3
	1
	0

+x4

	5
	4
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.8

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	−2	−7	2
	0	1	5	−2

x=x3

	−3
	−5
	1
	0

+x4

	2
	2
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.8

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	1	−4	−7	5
	0	1	2	−7

x=x3

	−1
	−2
	1
	0

+x4

	23
	7
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.9

Describe all solutions of

Ax=0

in parametric vector form, where A is row equivalent to the given matrix.

	4	−16	8
	−1	4	−2

x=x2

	4
	1
	0

+x3

	−2
	0
	1

(Type an integer or fraction for each matrix element.)

1.5.19

Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below.

3x1+3x2+6x3	=	12		3x1+3x2+6x3	=	0
−9x1−9x2−18x3	=	−36		−9x1−9x2−18x3	=	0
−4x2−4x3	=	16		−4x2−4x3	=	0

Describe the solution set,

x=

x1

x2

x3

,

of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or fraction for each matrix element.)

x=

x=x2  

x=

	8
	−4
	0

+x3

	−1
	−1
	1

  

x=x2+x3

Which option best compares the two systems?

The solution set of the first system is a plane parallel to the line that is the solution set of the second system.

The solution set of the first system is a plane parallel to the plane that is the solution set of the second system.

The solution set of the first system is a line parallel to the line that is the solution set of the second system.

The solution set of the first system is a line perpendicular to the line that is the solution set of the second system.

1.5.19

Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below.

2x1+2x2+4x3	=	8		2x1+2x2+4x3	=	0
−6x1−6x2−12x3	=	−24		−6x1−6x2−12x3	=	0
−7x2+14x3	=	14		−7x2+14x3	=	0

Describe the solution set,

x=

x1

x2

x3

,

of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or fraction for each matrix element.)

x=

x=x2  

x=

	6
	−2
	0

+x3

	−4
	2
	1

  

x=x2+x3

Which option best compares the two systems?

The solution set of the first system is a line parallel to the line that is the solution set of the second system.

The solution set of the first system is a line perpendicular to the line that is the solution set of the second system.

The solution set of the first system is a plane parallel to the line that is the solution set of the second system.

The solution set of the first system is a plane parallel to the plane that is the solution set of the second system.

1.5.19

Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below.

2x1+2x2+4x3	=	8		2x1+2x2+4x3	=	0
−4x1−4x2−8x3	=	−16		−4x1−4x2−8x3	=	0
−3x2+6x3	=	15		−3x2+6x3	=	0

Describe the solution set,

x=

x1

x2

x3

,

of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or fraction for each matrix element.)

x=

x=x2  

x=

	9
	−5
	0

+x3

	−4
	2
	1

  

x=x2+x3

Which option best compares the two systems?

The solution set of the first system is a line parallel to the line that is the solution set of the second system.

The solution set of the first system is a line perpendicular to the line that is the solution set of the second system.

The solution set of the first system is a plane parallel to the plane that is the solution set of the second system.

The solution set of the first system is a plane parallel to the line that is the solution set of the second system.

1.5.19

Describe the solutions of the first system of equations below in parametric vector form. Provide a geometric comparison with the solution set of the second system of equations below.

2x1+2x2+4x3	=	8		2x1+2x2+4x3	=	0
−4x1−4x2−8x3	=	−16		−4x1−4x2−8x3	=	0
−3x2+6x3	=	15		−3x2+6x3	=	0

Describe the solution set,

x=

x1

x2

x3

,

of the first system of equations in parametric vector form. Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or fraction for each matrix element.)

x=

x=x2  

x=

	9
	−5
	0

+x3

	−4
	2
	1

  

x=x2+x3

Which option best compares the two systems?

The solution set of the first system is a plane parallel to the plane that is the solution set of the second system.

The solution set of the first system is a line perpendicular to the line that is the solution set of the second system.

The solution set of the first system is a line parallel to the line that is the solution set of the second system.

The solution set of the first system is a plane parallel to the line that is the solution set of the second system.

1.5.28

Determine whether the statement below is true or false. Justify the answer.

If x is a nontrivial solution of

Ax=0,

then every entry in x is nonzero.

Choose the correct answer below.

The statement is false. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x cannot have any zero entries.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is false. The only solution of

Ax=0

is the zero vector. Thus, a nontrivial solution does not exist.

1.5.28

Determine whether the statement below is true or false. Justify the answer.

If x is a nontrivial solution of

Ax=0,

then every entry in x is nonzero.

Choose the correct answer below.

The statement is false. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is false. The only solution of

Ax=0

is the zero vector. Thus, a nontrivial solution does not exist.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x cannot have any zero entries.

1.5.28

Determine whether the statement below is true or false. Justify the answer.

If x is a nontrivial solution of

Ax=0,

then every entry in x is nonzero.

Choose the correct answer below.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x cannot have any zero entries.

The statement is false. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is true. A nontrivial solution of

Ax=0

is a nonzero vector x that satisfies

Ax=0.

Thus, a nontrivial solution x can have some zero entries so long as not all of its entries are zero.

The statement is false. The only solution of

Ax=0

is the zero vector. Thus, a nontrivial solution does not exist.

1.5.31

Determine whether the statement below is true or false. Justify the answer.

The homogenous equation

Ax=0

has the trivial solution if and only if the equation has at least one free variable.

Choose the correct answer below.

The statement is false. The homogeneous equation

Ax=0

always has the trivial solution.

The statement is true. The homogenous equation

Ax=0

has the trivial solution if and only if the equation has at least one free variable which implies that the equation has a nontrivial solution.

The statement is false. The homogeneous equation

Ax=0

never has the trivial solution.

The statement is true. The homogeneous equation

Ax=0

has the trivial solution if and only if the matrix A has a row of zeros which implies the equation has at least one free variable.

1.5.31

Determine whether the statement below is true or false. Justify the answer.

The homogenous equation

Ax=0

has the trivial solution if and only if the equation has at least one free variable.

Choose the correct answer below.

The statement is true. The homogenous equation

Ax=0

has the trivial solution if and only if the equation has at least one free variable which implies that the equation has a nontrivial solution.

The statement is false. The homogeneous equation

Ax=0

never has the trivial solution.

The statement is true. The homogeneous equation

Ax=0

has the trivial solution if and only if the matrix A has a row of zeros which implies the equation has at least one free variable.

The statement is false. The homogeneous equation

Ax=0

always has the trivial solution.

1.5.35

Determine whether the statement below is true or false. Justify the answer.

The solution set of

Ax=b

is the set of all vectors of the form

w=p+vh,

where

vh

is any solution of the equation

Ax=0.

Choose the correct answer below.

The statement is false. The solution set could be empty. The statement is only true when the equation

Ax=b

is consistent for some given

b,

and there exists a vector p such that p is a solution.

The statement is false. The solution set could be empty. The statement is only true when the equation

Ax=b

is inconsistent for some given

b,

and there exists a vector p such that p is a solution.

The statement is false. The solution set could be the trivial solution. The statement is only true when the equation

Ax=b

is inconsistent for some given

b,

and there exists a vector p such that p is a solution.

The statement is true. The equation

Ax=b

is always consistent and there always exists a vector p that is a solution.

1.5.35

Determine whether the statement below is true or false. Justify the answer.

The solution set of

Ax=b

is the set of all vectors of the form

w=p+vh,

where

vh

is any solution of the equation

Ax=0.

Choose the correct answer below.

The statement is true. The equation

Ax=b

is always consistent and there always exists a vector p that is a solution.

The statement is false. The solution set could be the trivial solution. The statement is only true when the equation

Ax=b

is inconsistent for some given

b,

and there exists a vector p such that p is a solution.

The statement is false. The solution set could be empty. The statement is only true when the equation

Ax=b

is consistent for some given

b,

and there exists a vector p such that p is a solution.

The statement is false. The solution set could be empty. The statement is only true when the equation

Ax=b

is inconsistent for some given

b,

and there exists a vector p such that p is a solution.

1.5.40

If

b≠0,

can the solution set of

Ax=b

be a plane through the origin?

Choose the correct answer.

Yes. Since the solution set of

Ax=0

contains the origin, the solution set of

Ax=b

must contain the origin.

Yes. The solution set of

Ax=b

is always represented as a plane through the origin.

No. The solution set of

Ax=b

contains the origin if and only if

Ax=b

is inconsistent, which is not true for any particular vector

b.

No. If the solution set of

Ax=b

contained the origin, then 0 would satisfy

A0=b,

which is not true since b is not the zero vector.

1.5.40

If

b≠0,

can the solution set of

Ax=b

be a plane through the origin?

Choose the correct answer.

No. The solution set of

Ax=b

contains the origin if and only if

Ax=b

is inconsistent, which is not true for any particular vector

b.

No. If the solution set of

Ax=b

contained the origin, then 0 would satisfy

A0=b,

which is not true since b is not the zero vector.

Yes. The solution set of

Ax=b

is always represented as a plane through the origin.

Yes. Since the solution set of

Ax=0

contains the origin, the solution set of

Ax=b

must contain the origin.

1.5.44

A is a

2×5

matrix with two pivot positions.

(a) Does the equation

Ax=0

have a nontrivial solution?

(b) Does the equation

Ax=b

have at least one solution for every possible

b?

(a) Does the equation

Ax=0

have a nontrivial solution?

Yes

No

(b) Does the equation

Ax=b

have at least one solution for every possible

b?

No

Yes

1.5.44

A is a

2×5

matrix with two pivot positions.

(a) Does the equation

Ax=0

have a nontrivial solution?

(b) Does the equation

Ax=b

have at least one solution for every possible

b?

(a) Does the equation

Ax=0

have a nontrivial solution?

Yes

No

(b) Does the equation

Ax=b

have at least one solution for every possible

b?

Yes

No

1.6.1

Suppose an economy has only two sectors: Goods and Services. Each year, Goods sells 75% of its outputs to Services and keeps the rest, while Services sells 66% of its output to Goods and retains the rest. Find equilibrium prices for the annual outputs of the Goods and Services sectors that make each sector's income match its expenditures.

0.34 0.75 0.66 0.25

Denote the prices (that is, dollar values) of the total annual outputs of the Goods and Services sectors by

pG

and

pS,

respectively.

If

pS=$1000,

then

pG=$880.

(Type an integer or a decimal.)

If

pS=$75,

then

pG=$66.

(Type an integer or a decimal.)

1.6.1

Suppose an economy has only two sectors: Goods and Services. Each year, Goods sells 80% of its outputs to Services and keeps the rest, while Services sells 65% of its output to Goods and retains the rest. Find equilibrium prices for the annual outputs of the Goods and Services sectors that make each sector's income match its expenditures.

0.35 0.8 0.65 0.2

Denote the prices (that is, dollar values) of the total annual outputs of the Goods and Services sectors by

pG

and

pS,

respectively.

If

pS=$1000,

then

pG=$812.5.

(Type an integer or a decimal.)

If

pS=$80,

then

pG=$65.

(Type an integer or a decimal.)

1.6.1

Suppose an economy has only two sectors: Goods and Services. Each year, Goods sells 75% of its outputs to Services and keeps the rest, while Services sells 60% of its output to Goods and retains the rest. Find equilibrium prices for the annual outputs of the Goods and Services sectors that make each sector's income match its expenditures.

0.4 0.75 0.6 0.25

Denote the prices (that is, dollar values) of the total annual outputs of the Goods and Services sectors by

pG

and

pS,

respectively.

If

pS=$1000,

then

pG=$800.

(Type an integer or a decimal.)

If

pS=$75,

then

pG=$60.

(Type an integer or a decimal.)

1.6.2

Suppose an economy consists of the Coal, Electric, and Steel sectors. Denote the prices (that is, dollar values) of the total annual outputs of the Coal, Electric, and Steel sectors by

pC,

pE,

and

pS,

respectively. Suppose the general solution to find equilibrium prices that make each sector's income match its expenditures is

pC=0.91pS,

pE=0.86pS,

and

pS

is free. One set of equilibrium prices for this economy is

pC=$91,

pE=$86,

and

pS=$100.

Find another set. Suppose the same economy used Japanese yen instead of dollars to measure the values of the various sector's output. Would this change the problem in any way? Discuss.

If

pS=$200,

then

pC=$182

and

pE=$172.

(Type integers or decimals.)

How would changing the unit of measurement to Japanese yen change this problem?

It has the same effect as multiplying all equilibrium prices by a constant. The ratios of the prices will change, but the prices will remain the same.

It has the same effect as multiplying all equilibrium prices by a constant. The prices will change, but the ratios of the prices will remain the same.

It has the same effect as multiplying all equilibrium prices by their reciprocal and some constant. The prices will change, but the ratios of the prices will remain the same.

It has the same effect as multiplying all equilibrium prices by their reciprocal and some constant. The prices and the ratios of the prices will change.

1.6.2

Suppose an economy consists of the Coal, Electric, and Steel sectors. Denote the prices (that is, dollar values) of the total annual outputs of the Coal, Electric, and Steel sectors by

pC,

pE,

and

pS,

respectively. Suppose the general solution to find equilibrium prices that make each sector's income match its expenditures is

pC=0.87pS,

pE=0.89pS,

and

pS

is free. One set of equilibrium prices for this economy is

pC=$87,

pE=$89,

and

pS=$100.

Find another set. Suppose the same economy used Japanese yen instead of dollars to measure the values of the various sector's output. Would this change the problem in any way? Discuss.

If

pS=$200,

then

pC=$174

and

pE=$178.

(Type integers or decimals.)

How would changing the unit of measurement to Japanese yen change this problem?

It has the same effect as multiplying all equilibrium prices by their reciprocal and some constant. The prices will change, but the ratios of the prices will remain the same.

It has the same effect as multiplying all equilibrium prices by a constant. The prices will change, but the ratios of the prices will remain the same.

It has the same effect as multiplying all equilibrium prices by their reciprocal and some constant. The prices and the ratios of the prices will change.

It has the same effect as multiplying all equilibrium prices by a constant. The ratios of the prices will change, but the prices will remain the same.

1.6.3

Consider an economy with three sectors, Chemicals & Metals, Fuels & Power, and Machinery. Chemicals sells

30%

of its output to Fuels and

60%

to Machinery and retains the rest. Fuels sells

80%

of its output to Chemicals and

10%

to Machinery and retains the rest. Machinery sells

30%

of its output to Chemicals and

20%

to Fuels and retains the rest. Complete parts (a) through (c) below.

a. Construct the exchange table for this economy.

Distribution of Output from:
Chemicals	Fuels	Machinery	Purchased by:
0.10	0.80	0.30	Chemicals
0.30	0.10	0.20	Fuels
0.60	0.10	0.50	Machinery

(Type integers or decimals.)

b. Develop a system of equations that leads to prices at which each sector's income matches its expenses. Then write the augmented matrix that can be row reduced to find these prices. The first, second, and third columns of the matrix should correspond to Chemicals, Fuels, and Machinery, respectively.

The augmented matrix is

	0.90	−0.80	−0.30	0
	−0.30	0.90	−0.20	0
	−0.60	−0.10	0.50	0

.

(Type an integer or decimal for each matrix element.)

c. Find a set of equilibrium prices when the price for the Machinery output is

20

units.

pChemicals=15.1,

pFuels=9.5,

pMachinery=20

(Type integers or decimals rounded to the nearest tenth as needed.)

1.6.3

Consider an economy with three sectors, Chemicals & Metals, Fuels & Power, and Machinery. Chemicals sells

20%

of its output to Fuels and

60%

to Machinery and retains the rest. Fuels sells

70%

of its output to Chemicals and

20%

to Machinery and retains the rest. Machinery sells

50%

of its output to Chemicals and

30%

to Fuels and retains the rest. Complete parts (a) through (c) below.

a. Construct the exchange table for this economy.

Distribution of Output from:
Chemicals	Fuels	Machinery	Purchased by:
0.20	0.70	0.50	Chemicals
0.20	0.10	0.30	Fuels
0.60	0.20	0.20	Machinery

(Type integers or decimals.)

b. Develop a system of equations that leads to prices at which each sector's income matches its expenses. Then write the augmented matrix that can be row reduced to find these prices. The first, second, and third columns of the matrix should correspond to Chemicals, Fuels, and Machinery, respectively.

The augmented matrix is

	0.8	−0.7	−0.5	0
	−0.2	0.9	−0.3	0
	−0.6	−0.2	0.8	0

.

(Type an integer or decimal for each matrix element.)

c. Find a set of equilibrium prices when the price for the Machinery output is

90

units.

pChemicals=102.4,

pFuels=52.8,

pMachinery=90.0

(Type integers or decimals rounded to the nearest tenth as needed.)

1.6.4

Suppose an economy has four sectors: Mining, Lumber, Energy, and Transportation. Mining sells

15%

of its output to Lumber,

70%

to Energy, and retains the rest. Lumber sells

15%

of its output to Mining,

50%

to Energy,

20%

to Transportation, and retains the rest. Energy sells

30%

of its output to Mining,

10%

to Lumber,

25%

to Transportation, and retains the rest. Transportation sells

10%

of its output to Mining,

20%

to Lumber,

50%

to Energy, and retains the rest.

a. Construct the exchange table for this economy.

b. Find a set of equilibrium prices for this economy.

a. Complete the exchange table below.

Distribution of Output from:
Mining	Lumber	Energy	Transportation	Purchased by:
0.15	0.15	0.30	0.10	Mining
0.15	0.15	0.10	0.20	Lumber
0.70	0.50	0.35	0.50	Energy
0.00	0.20	0.25	0.20	Transportation

(Type integers or decimals.)

b. Denote the prices (that is, dollar values) of the total annual outputs of the Mining, Lumber, Energy, and Transportation sectors by

pM,

pL,

pE,

and

pT,

respectively.

If

pT=$100,

then

pM=$117,

pL=$75,

and

pE=$260.

(Round to the nearest dollar as needed.)

1.6.4

Suppose an economy has four sectors: Mining, Lumber, Energy, and Transportation. Mining sells

10%

of its output to Lumber,

50%

to Energy, and retains the rest. Lumber sells

10%

of its output to Mining,

40%

to Energy,

25%

to Transportation, and retains the rest. Energy sells

30%

of its output to Mining,

10%

to Lumber,

20%

to Transportation, and retains the rest. Transportation sells

25%

of its output to Mining,

10%

to Lumber,

50%

to Energy, and retains the rest.

a. Construct the exchange table for this economy.

b. Find a set of equilibrium prices for this economy.

a. Complete the exchange table below.

Distribution of Output from:
Mining	Lumber	Energy	Transportation	Purchased by:
0.40	0.10	0.30	0.25	Mining
0.10	0.25	0.10	0.10	Lumber
0.50	0.40	0.40	0.50	Energy
0.00	0.25	0.20	0.15	Transportation

(Type integers or decimals.)

b. Denote the prices (that is, dollar values) of the total annual outputs of the Mining, Lumber, Energy, and Transportation sectors by

pM,

pL,

pE,

and

pT,

respectively.

If

pT=$100,

then

pM=$215,

pL=$85,

and

pE=$319.

(Round to the nearest dollar as needed.)

1.6.5

Balance the chemical equation.

CS2+NH3→H2S+NH4SCN

Assume the coefficient of

NH4SCN

is

1.

What is the balanced equation?

1CS2+2NH3→1H2S+1NH4SCN

1.6.5

Balance the chemical equation.

B2S3+H2O→H3BO3+H2S

Assume the coefficient of

H2S

is

3.

What is the balanced equation?

1B2S3+6H2O→2H3BO3+3H2S

1.6.5

Balance the chemical equation.

B2S3+H2O→H3BO3+H2S

Assume the coefficient of

H2S

is

3.

What is the balanced equation?

1B2S3+6H2O→2H3BO3+3H2S

1.6.5

Balance the chemical equation.

B2S3+H2O→H3BO3+H2S

Assume the coefficient of

H2S

is

3.

What is the balanced equation?

1B2S3+6H2O→2H3BO3+3H2S

1.6.7

Balance the following chemical equation.

NaHCO3+H3C6H5O7→Na3C6H5O7+H2O+CO2

Assume the coefficient of

CO2

is

3.

What is the balanced equation?

3NaHCO3+1H3C6H5O7→1Na3C6H5O7+3H2O+3CO2

1.6.9

Balance the chemical equation.

PbN6+CrMn2O8→Pb3O4+Cr2O3+MnO2+NO

Assume the coefficient of

NO

is

90.

What is the balanced equation?

15PbN6+44CrMn2O8→5Pb3O4+22Cr2O3+88MnO2+90NO

1.6.9

Balance the chemical equation.

Au+HCl+HNO3→AuCl3+NO+H2O

Assume the coefficient of

H2O

is

2.

What is the balanced equation?

1Au+3HCl+1HNO3→1AuCl3+1NO+2H2O

1.6.9

Balance the chemical equation.

PbN6+CrMn2O8→Pb3O4+Cr2O3+MnO2+NO

Assume the coefficient of

NO

is

90.

What is the balanced equation?

15PbN6+44CrMn2O8→5Pb3O4+22Cr2O3+88MnO2+90NO

1.6.9

Balance the chemical equation.

Au+HCl+HNO3→AuCl3+NO+H2O

Assume the coefficient of

H2O

is

2.

What is the balanced equation?

1Au+3HCl+1HNO3→1AuCl3+1NO+2H2O

1.7.5

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

0 −8 16 3 1 −14 −1 5 −4 1 −5 −2

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

	0	−8	16	0
	3	1	−14	0
	−1	5	−4	0
	1	−5	−2	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

1.7.5

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

0 −3 9 2 1 −7 −1 4 −8 1 −4 −2

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

	0	−3	9	0
	2	1	−7	0
	−1	4	−8	0
	1	−4	−2	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

1.7.5

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

0 −8 16 3 1 −14 −1 5 −6 1 −5 −2

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

	0	−8	16	0
	3	1	−14	0
	−1	5	−6	0
	1	−5	−2	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

1.7.5

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

0 −3 9 2 1 −7 −1 4 −6 1 −4 −2

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

	0	−3	9	0
	2	1	−7	0
	−1	4	−6	0
	1	−4	−2	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

1.7.5

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

0 −3 9 2 1 −7 −1 4 −7 1 −4 −2

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

	0	−3	9	0
	2	1	−7	0
	−1	4	−7	0
	1	−4	−2	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

1.7.6

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

−4 −3 0 0 −1 6 1 1 −6 2 1 −12

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

	−4	−3	0	0
	0	−1	6	0
	1	1	−6	0
	2	1	−12	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

1.7.6

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

−4 −3 0 0 −1 5 1 1 −5 2 1 −10

Select the correct choice below and fill in the answer box within your choice.

(Type an integer or simplified fraction for each matrix element.)

If A is the given matrix, then the augmented matrix

	−4	−3	0	0
	0	−1	5	0
	1	1	−5	0
	2	1	−10	0

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has more than one solution. Therefore, the columns of A do not form a linearly independent set.

If A is the given matrix, then the augmented matrix

represents the equation

Ax=0.

The reduced echelon form of this matrix indicates that

Ax=0

has only the trivial solution. Therefore, the columns of A do not form a linearly independent set.

1.7.8

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

1	−2	3
−2	4	3

Choose the correct answer below.

The columns of the matrix do not form a linearly independent set because the set contains more vectors than there are entries in each vector.

The columns of the matrix do form a linearly independent set because there are more entries in each vector than there are vectors in the set.

The columns of the matrix do form a linearly independent set because the set contains more vectors than there are entries in each vector.

The columns of the matrix do not form a linearly independent set because there are more entries in each vector than there are vectors in the set.

1.7.8

Determine if the columns of the matrix form a linearly independent set. Justify your answer.

1	−2	3	4
−2	4	−6	4
0	1	−1	5

Choose the correct answer below.

The columns of the matrix do not form a linearly independent set because the set contains more vectors than there are entries in each vector.

The columns of the matrix do not form a linearly independent set because there are more entries in each vector than there are vectors in the set.

The columns of the matrix do form a linearly independent set because the set contains more vectors than there are entries in each vector.

The columns of the matrix do form a linearly independent set because there are more entries in each vector than there are vectors in the set.

1.7.13

Find the value(s) of h for which the vectors are linearly dependent.

1

3

−2

,

−4

−11

8

,

3

h

−6

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The vectors are linearly dependent if

h=

because the related matrix will have a free variable. (Type an integer or a simplified fraction.)

The vectors are linearly dependent for all values of h because the related matrix always has a free variable.

The vectors are linearly independent for all values of h because the related matrix never has a free variable.

1.7.13

Find the value(s) of h for which the vectors are linearly dependent.

1

5

−2

,

−2

−9

4

,

2

h

−4

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The vectors are linearly dependent if

h=

because the related matrix will have a free variable. (Type an integer or a simplified fraction.)

The vectors are linearly dependent for all values of h because the related matrix always has a free variable.

The vectors are linearly independent for all values of h because the related matrix never has a free variable.

1.7.14

Find the value(s) of h for which the vectors are linearly dependent. Justify your answer.

1

−3

−6

,

−4

13

8

,

4

1

h

The value(s) of h which makes the vectors linearly dependent is(are)

−232

because this will cause

x3

to be a

free

variable.

(Use a comma to separate answers as needed.)

1.7.14

Find the value(s) of h for which the vectors are linearly dependent. Justify your answer.

1

−2

−4

,

−3

7

6

,

3

1

h

The value(s) of h which makes the vectors linearly dependent is(are)

−54

because this will cause

x3

to be a

free

variable.

(Use a comma to separate answers as needed.)

1.7.16

Determine by inspection whether the vectors are linearly independent. Justify your answer.

6

−12

9

,

4

−8

6

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The set of vectors is linearly dependent because

23

times the first vector is equal to the second vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly dependent because neither vector is the zero vector.

The set of vectors is linearly independent because neither vector is a multiple of the other vector.

1.7.16

Determine by inspection whether the vectors are linearly independent. Justify your answer.

6

−4

2

,

9

−6

3

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The set of vectors is linearly dependent because

32

times the first vector is equal to the second vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly dependent because neither vector is the zero vector.

The set of vectors is linearly independent because neither vector is a multiple of the other vector.

1.7.17

Determine by inspection whether the vectors are linearly independent. Justify your answer.

4

2

−3

,

0

0

0

,

−2

4

4

Choose the correct answer below.

The set of vectors is linearly independent because

times the first vector is equal to the second vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly dependent because

times the first vector is equal to the third vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly dependent because one of the vectors is the zero vector.

The set of vectors is linearly independent because none of the vectors are multiples of the other vectors.

1.7.17

Determine by inspection whether the vectors are linearly independent. Justify your answer.

6

2

−3

,

0

0

0

,

−4

3

2

Choose the correct answer below.

The set of vectors is linearly dependent because

times the first vector is equal to the third vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly independent because

times the first vector is equal to the second vector.

(Type an integer or a simplified fraction.)

The set of vectors is linearly dependent because one of the vectors is the zero vector.

The set of vectors is linearly independent because none of the vectors are multiples of the other vectors.

1.7.19

Determine by inspection whether the vectors are linearly independent. Justify your answer.

−8

16

4

,

2

−4

1

Choose the correct answer below.

The set is linearly independent because the first vector is a multiple of the other vector. The entries in the first vector are

−4

times the corresponding entry in the second vector.

The set is linearly dependent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−4

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

The set is linearly independent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−4

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

The set is linearly dependent because the first vector is a multiple of the other vector. The entries in the first vector are

−4

times the corresponding entry in the second vector.

1.7.19

Determine by inspection whether the vectors are linearly independent. Justify your answer.

10

−20

−5

,

−2

4

−1

Choose the correct answer below.

The set is linearly dependent because the first vector is a multiple of the other vector. The entries in the first vector are

−5

times the corresponding entry in the second vector.

The set is linearly independent because the first vector is a multiple of the other vector. The entries in the first vector are

−5

times the corresponding entry in the second vector.

The set is linearly independent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−5

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

The set is linearly dependent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−5

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

1.7.19

Determine by inspection whether the vectors are linearly independent. Justify your answer.

8

−12

−4

,

−2

3

−1

Choose the correct answer below.

The set is linearly independent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−4

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

The set is linearly independent because the first vector is a multiple of the other vector. The entries in the first vector are

−4

times the corresponding entry in the second vector.

The set is linearly dependent because the first vector is a multiple of the other vector. The entries in the first vector are

−4

times the corresponding entry in the second vector.

The set is linearly dependent because neither vector is a multiple of the other vector. Two of the entries in the first vector are

−4

times the corresponding entry in the second vector. But this multiple does not work for the third entries.

1.7.23

Determine whether the statement below is true or false. Justify the answer.

If S is a linearly dependent set, then each vector is a linear combination of the other vectors in S.

Choose the correct answer below.

The statement is false. If an indexed set of vectors, S, is linearly dependent, then it is only necessary that one of the vectors is a linear combination of the other vectors in the set.

The statement is false. If S is linearly dependent then there is at least one vector that is not a linear combination of the other vectors, but the others may be linear combinations of each other.

The statement is true. If an indexed set of vectors, S, is linearly dependent, then at least one of the vectors can be written as a linear combination of other vectors in the set. Using the basic properties of equality, each of the vectors in the linear combination can also be written as a linear combination of those vectors.

The statement is true. If S is linearly dependent then for each j,

vj,

a vector in S, is a linear combination of the preceding vectors in S.

1.7.23

Determine whether the statement below is true or false. Justify the answer.

If S is a linearly dependent set, then each vector is a linear combination of the other vectors in S.

Choose the correct answer below.

The statement is false. If S is linearly dependent then there is at least one vector that is not a linear combination of the other vectors, but the others may be linear combinations of each other.

The statement is true. If an indexed set of vectors, S, is linearly dependent, then at least one of the vectors can be written as a linear combination of other vectors in the set. Using the basic properties of equality, each of the vectors in the linear combination can also be written as a linear combination of those vectors.

The statement is false. If an indexed set of vectors, S, is linearly dependent, then it is only necessary that one of the vectors is a linear combination of the other vectors in the set.

The statement is true. If S is linearly dependent then for each j,

vj,

a vector in S, is a linear combination of the preceding vectors in S.

1.7.24

Determine whether the statement below is true or false. Justify the answer.

If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.

Choose the correct answer below.

The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.

The statement is false. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. If there are fewer vectors than entries in the vectors, then at least one of the vectors must be written as a linear combination of the others.

The statement is true. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly independent. One example is a set consisting of two vectors where one of the vectors is not a scalar multiple of the other vector.

The statement is true. If a set contains fewer vectors than there are entries in the vectors, then there are less variables than equations, so there cannot be any free variables in the equation

Ax=0.

1.7.24

Determine whether the statement below is true or false. Justify the answer.

If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.

Choose the correct answer below.

The statement is true. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly independent. One example is a set consisting of two vectors where one of the vectors is not a scalar multiple of the other vector.

The statement is false. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. If there are fewer vectors than entries in the vectors, then at least one of the vectors must be written as a linear combination of the others.

The statement is true. If a set contains fewer vectors than there are entries in the vectors, then there are less variables than equations, so there cannot be any free variables in the equation

Ax=0.

The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.

1.7.24

Determine whether the statement below is true or false. Justify the answer.

If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.

Choose the correct answer below.

The statement is true. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly independent. One example is a set consisting of two vectors where one of the vectors is not a scalar multiple of the other vector.

The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.

The statement is false. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. If there are fewer vectors than entries in the vectors, then at least one of the vectors must be written as a linear combination of the others.

The statement is true. If a set contains fewer vectors than there are entries in the vectors, then there are less variables than equations, so there cannot be any free variables in the equation

Ax=0.

1.7.24

Determine whether the statement below is true or false. Justify the answer.

If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.

Choose the correct answer below.

The statement is true. If a set contains fewer vectors than there are entries in the vectors, then there are less variables than equations, so there cannot be any free variables in the equation

Ax=0.

The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.

The statement is true. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly independent. One example is a set consisting of two vectors where one of the vectors is not a scalar multiple of the other vector.

The statement is false. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. If there are fewer vectors than entries in the vectors, then at least one of the vectors must be written as a linear combination of the others.

1.7.24

Determine whether the statement below is true or false. Justify the answer.

If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.

Choose the correct answer below.

The statement is false. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly dependent. One example is a set consisting of two vectors where one of the vectors is a scalar multiple of the other vector.

The statement is true. There exists a set that contains fewer vectors than there are entries in the vectors that is linearly independent. One example is a set consisting of two vectors where one of the vectors is not a scalar multiple of the other vector.

The statement is false. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. If there are fewer vectors than entries in the vectors, then at least one of the vectors must be written as a linear combination of the others.

The statement is true. If a set contains fewer vectors than there are entries in the vectors, then there are less variables than equations, so there cannot be any free variables in the equation

Ax=0.

1.7.25

Determine whether the statement below is true or false. Justify the answer.

The columns of any

4×5

matrix are linearly dependent.

Choose the correct answer below.

The statement is false. If a matrix has more rows than columns then the columns of the matrix are linearly dependent.

The statement is true. When a

4×5

matrix is written in reduced echelon form, there will be at least one row of zeros, so the columns of the matrix are linearly dependent.

The statement is true. A

4×5

matrix has more columns than rows, and if a set contains more vectors than there are entries in each vector, then the set is linearly dependent.

The statement is false. If A is a

4×5

matrix then the matrix equation

Ax=0

is inconsistent because the reduced echelon augmented matrix has a row with all zeros except in the last column.

1.7.25

Determine whether the statement below is true or false. Justify the answer.

The columns of any

4×5

matrix are linearly dependent.

Choose the correct answer below.

The statement is true. When a

4×5

matrix is written in reduced echelon form, there will be at least one row of zeros, so the columns of the matrix are linearly dependent.

The statement is false. If a matrix has more rows than columns then the columns of the matrix are linearly dependent.

The statement is true. A

4×5

matrix has more columns than rows, and if a set contains more vectors than there are entries in each vector, then the set is linearly dependent.

The statement is false. If A is a

4×5

matrix then the matrix equation

Ax=0

is inconsistent because the reduced echelon augmented matrix has a row with all zeros except in the last column.

1.7.32

Describe the possible echelon forms of the following matrix.

A is a

3×3

matrix,

A=

a1

a2

a3

,

such that

{a1,

a2}

is linearly independent and

a3

is not in

Span{a1,

a2}.

Select all that apply. (Note that leading entries marked with an X may have any nonzero value and starred entries

(*)

may have any value including zero.)

X	0	0
0	0	0
0	0	0

0	X	*
0	0	X
0	0	0

0	0	X
0	0	0
0	0	0

X	*	*
0	X	*
0	0	X

1.7.32

Describe the possible echelon forms of the following matrix.

A is a

4×3

matrix,

A=

a1

a2

a3

,

such that

{a1,

a2}

is linearly independent and

a3

is not in

Span{a1,

a2}.

Select all that apply. (Note that leading entries marked with an X may have any nonzero value and starred entries

(*)

may have any value including zero.)

0	X	*
0	0	X
0	0	0
0	0	0

X	*	*
X	*	*
X	*	*
0	X	X

X	*	*
0	X	*
0	0	X
0	0	0

X	0	0
0	0	0
0	0	0
0	0	0

1.7.42

The following statement is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If the statement is true, give a justification.

If

v1,

v2,

v3

are in

ℝ3

and

v3

is not a linear combination of

v1,

v2,

then

{v1,

v2,

v3}

is linearly independent.

Fill in the blanks below.

The statement is

false.

Take

v1

and

v2

to be multiples of one vector and take

v3

to be not a multiple of that vector. For example,

v1=

1

1

1

,

v2=

2

2

2

,

v3=

1

0

0

.

Since at least one of the vectors is a linear combination of the other two, the three vectors are linearly

dependent.

1.7.42

The following statement is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If the statement is true, give a justification.

If

v1,

v2,

v3

are in

ℝ3

and

v3

is not a linear combination of

v1,

v2,

then

{v1,

v2,

v3}

is linearly independent.

Fill in the blanks below.

The statement is

false.

Take

v1

and

v2

to be multiples of one vector and take

v3

to be not a multiple of that vector. For example,

v1=

1

1

1

,

v2=

2

2

2

,

v3=

1

0

0

.

Since at least one of the vectors is a linear combination of the other two, the three vectors are linearly

dependent.

1.8.1

Let

A=

	2	0
	0	2

,

and define

T : ℝ2→ℝ2

by

T(x)=Ax.

Find the images under T of

u=

	2
	−6

and

v=

	a
	b

.

T(u)=

	4
	−12

T(v)=

	2a
	2b

1.8.1

Let

A=

	3	0
	0	3

,

and define

T : ℝ2→ℝ2

by

T(x)=Ax.

Find the images under T of

u=

	4
	−5

and

v=

	a
	b

.

T(u)=

	12
	−15

T(v)=

	3a
	3b

1.8.4

If T is defined by

T(x)=Ax,

find a vector x whose image under T is

b,

and determine whether x is unique. Let

A=

	1	−3	4
	0	1	−5
	4	−13	16

and

b=

	−4
	−8
	−3

.

Find a single vector x whose image under T is

b.

x=

	−39
	−13
	−1

Is the vector x found in the previous step unique?

No, because there are no free variables in the system of equations.

Yes, because there is a free variable in the system of equations.

Yes, because there are no free variables in the system of equations.

No, because there is a free variable in the system of equations.

1.8.4

If T is defined by

T(x)=Ax,

find a vector x whose image under T is

b,

and determine whether x is unique. Let

A=

	1	−4	4
	0	1	−4
	2	−9	8

and

b=

	−4
	−1
	−3

.

Find a single vector x whose image under T is

b.

x=

	−20
	−5
	−1

Is the vector x found in the previous step unique?

Yes, because there are no free variables in the system of equations.

No, because there are no free variables in the system of equations.

Yes, because there is a free variable in the system of equations.

No, because there is a free variable in the system of equations.

1.8.4

If T is defined by

T(x)=Ax,

find a vector x whose image under T is

b,

and determine whether x is unique. Let

A=

	1	−2	3
	0	1	−3
	3	−7	9

and

b=

	−4
	−6
	−3

.

Find a single vector x whose image under T is

b.

x=

	−19
	−9
	−1

Is the vector x found in the previous step unique?

No, because there is a free variable in the system of equations.

Yes, because there are no free variables in the system of equations.

Yes, because there is a free variable in the system of equations.

No, because there are no free variables in the system of equations.

1.8.5

If T is defined by

T(x)=Ax,

find a vector x whose image under T is

b,

and determine whether x is unique. Let

A=

	1	−7	−19
	−4	21	55

and

b=

	−3
	5

.

Find a single vector x whose image under T is

b.

x=

	4
	1
	0

Is the vector x found in the previous step unique?

Yes, because there are no free variables in the system of equations.

No, because there are no free variables in the system of equations.

Yes, because there is a free variable in the system of equations.

No, because there is a free variable in the system of equations.

1.8.5

If T is defined by

T(x)=Ax,

find a vector x whose image under T is

b,

and determine whether x is unique. Let

A=

	1	−3	−10
	−4	6	16

and

b=

	−1
	−2

.

Find a single vector x whose image under T is

b.

x=

	2
	1
	0

Is the vector x found in the previous step unique?

No, because there are no free variables in the system of equations.

No, because there is a free variable in the system of equations.

Yes, because there is a free variable in the system of equations.

Yes, because there are no free variables in the system of equations.

1.8.6

Find a vector x whose image under T, defined by

T(x)=Ax,

is

b,

and determine whether x is unique. Let

A=

	1	3	6
	4	15	27
	0	1	1
	−4	−13	−25

,

b=

	16
	73
	3
	−67

.

Find a single vector x whose image under T is

b.

x=

	7
	3
	0

Is the vector x found in the previous step unique?

No, because there are no free variables in the system of equations.

Yes, because there are no free variables in the system of equations.

Yes, because there is a free variable in the system of equations.

No, because there is a free variable in the system of equations.

1.8.6

Find a vector x whose image under T, defined by

T(x)=Ax,

is

b,

and determine whether x is unique. Let

A=

	1	3	7
	2	9	17
	0	1	1
	−2	−7	−15

,

b=

	12
	30
	2
	−26

.

Find a single vector x whose image under T is

b.

x=

	6
	2
	0

Is the vector x found in the previous step unique?

Yes, because there is a free variable in the system of equations.

Yes, because there are no free variables in the system of equations.

No, because there is a free variable in the system of equations.

No, because there are no free variables in the system of equations.

1.8.7

Let A be a

4×6

matrix. What must a and b be in order to define

T : ℝa→ℝb

by

T(x)=Ax?

a=6

(Simplify your answer.)

b=4

(Simplify your answer.)

1.8.7

Let A be a

7×4

matrix. What must a and b be in order to define

T : ℝa→ℝb

by

T(x)=Ax?

a=4

(Simplify your answer.)

b=7

(Simplify your answer.)

1.8.10

Find all x in

ℝ4

that are mapped into the zero vector by the transformation

x↦Ax

for the given matrix A.

A=

	1	2	10	1
	1	0	2	−4
	0	1	4	5
	−1	7	26	24

Select the correct choice below and fill in the answer box within your choice.

There is only one vector, which is

x=.

x3

	−2
	−4
	1
	0

x1+x3+x4

x1+x2

1.8.10

Find all x in

ℝ4

that are mapped into the zero vector by the transformation

x↦Ax

for the given matrix A.

A=

	1	2	6	−1
	1	0	2	−4
	0	1	2	3
	−2	4	4	10

Select the correct choice below and fill in the answer box within your choice.

There is only one vector, which is

x=.

x1+x3+x4

x1+x2

x3

	−2
	−2
	1
	0

1.8.10

Find all x in

ℝ4

that are mapped into the zero vector by the transformation

x↦Ax

for the given matrix A.

A=

	1	2	6	−1
	1	0	2	−4
	0	1	2	3
	−3	1	−4	8

Select the correct choice below and fill in the answer box within your choice.

There is only one vector, which is

x=.

x3

	−2
	−2
	1
	0

x1+x3+x4

x1+x2

1.8.10

Find all x in

ℝ4

that are mapped into the zero vector by the transformation

x↦Ax

for the given matrix A.

A=

	1	3	13	3
	1	0	4	−5
	0	1	3	4
	−1	6	14	13

Select the correct choice below and fill in the answer box within your choice.

There is only one vector, which is

x=.

x3

	−4
	−3
	1
	0

x1+x3+x4

x1+x2

1.8.10

Find all x in

ℝ4

that are mapped into the zero vector by the transformation

x↦Ax

for the given matrix A.

A=

	1	4	20	11
	1	0	4	−4
	0	1	4	5
	−2	1	−4	5

Select the correct choice below and fill in the answer box within your choice.

There is only one vector, which is

x=.

x1+x3+x4

x3

	−4
	−4
	1
	0

x1+x2

1.8.12

Let

b=

	−13
	1
	−3
	19

,

and let A be the matrix

	1	4	12	5
	1	0	4	−5
	0	1	2	3
	−3	−4	−20	−11

.

Is b in the range of the linear transformation

x↦Ax?

Why or why not?

Is b in the range of the linear transformation? Why or why not?

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

1.8.12

Let

b=

	−3
	2
	−2
	3

,

and let A be the matrix

	1	2	7	0
	1	0	3	−5
	0	1	2	3
	−4	−1	−14	1

.

Is b in the range of the linear transformation

x↦Ax?

Why or why not?

Is b in the range of the linear transformation? Why or why not?

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

1.8.12

Let

b=

	−13
	1
	−3
	19

,

and let A be the matrix

	1	4	20	15
	1	0	4	−3
	0	1	4	5
	−3	−4	−28	−21

.

Is b in the range of the linear transformation

x↦Ax?

Why or why not?

Is b in the range of the linear transformation? Why or why not?

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

No, b is not in the range of the linear transformation because the system represented by the appropriate augmented matrix is consistent.

Yes, b is in the range of the linear transformation because the system represented by the appropriate augmented matrix is inconsistent.

1.8.17

Let T:

ℝ2→ℝ2

be a linear transformation that maps

u=

3

5

into

4

1

and maps

v=

3

3

into

−1

3

.

Use the fact that T is linear to find the images under T of

3u,

4v,

and 3u+4v.

The image of

3u

is

	12
	3

.

The image of

4v

is

	−4
	12

.

The image of

3u+4v

is

	8
	15

.

1.8.17

Let T:

ℝ2→ℝ2

be a linear transformation that maps

u=

3

4

into

6

1

and maps

v=

3

3

into

−1

3

.

Use the fact that T is linear to find the images under T of

2u,

4v,

and 2u+4v.

The image of

2u

is

	12
	2

.

The image of

4v

is

	−4
	12

.

The image of

2u+4v

is

	8
	14

.

1.8.20

Let

x=

	x1
	x2

,

v1=

	8
	−2

,

and

v2=

	9
	−6

,

and let T :

ℝ2→ℝ2

be a linear transformation that maps x into

x1v1+x2v2.

Find a matrix A such that

T(x)

is

Ax

for each

x.

A=

	8	9
	−2	−6

1.8.20

Let

x=

	x1
	x2

,

v1=

	−6
	5

,

and

v2=

	4
	−7

,

and let T :

ℝ2→ℝ2

be a linear transformation that maps x into

x1v1+x2v2.

Find a matrix A such that

T(x)

is

Ax

for each

x.

A=

	−6	4
	5	−7

1.8.21

Determine whether the statement below is true or false. Justify the answer.

A linear transformation is a special type of function.

Choose the correct answer below.

The statement is false. A linear transformation is not a function because it maps one vector x to more than one vector

T(x).

The statement is true. A linear transformation is a function from

ℝ

to

ℝ

that assigns to each vector x in

ℝ

a vector

T(x)

in

ℝ.

The statement is false. A linear transformation is not a function because it maps more than one vector x to the same vector

T(x).

The statement is true. A linear transformation is a function from

ℝn

to

ℝm

that assigns to each vector x in

ℝn

a vector

T(x)

in

ℝm.

1.8.21

Determine whether the statement below is true or false. Justify the answer.

A linear transformation is a special type of function.

Choose the correct answer below.

The statement is false. A linear transformation is not a function because it maps more than one vector x to the same vector

T(x).

The statement is false. A linear transformation is not a function because it maps one vector x to more than one vector

T(x).

The statement is true. A linear transformation is a function from

ℝ

to

ℝ

that assigns to each vector x in

ℝ

a vector

T(x)

in

ℝ.

The statement is true. A linear transformation is a function from

ℝn

to

ℝm

that assigns to each vector x in

ℝn

a vector

T(x)

in

ℝm.

1.8.22

Determine whether the statement below is true or false. Justify the answer.

Every matrix transformation is a linear transformation.

Choose the correct answer below.

The statement is false. Not every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the property

T(u+v)=T(u)+T(v),

but not all matrix transformations have the property

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is false. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

1.8.22

Determine whether the statement below is true or false. Justify the answer.

Every matrix transformation is a linear transformation.

Choose the correct answer below.

The statement is false. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the property

T(u+v)=T(u)+T(v),

but not all matrix transformations have the property

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is false. Not every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

1.8.22

Determine whether the statement below is true or false. Justify the answer.

Every matrix transformation is a linear transformation.

Choose the correct answer below.

The statement is false. Not every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is false. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the property

T(u+v)=T(u)+T(v),

but not all matrix transformations have the property

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

1.8.22

Determine whether the statement below is true or false. Justify the answer.

Every matrix transformation is a linear transformation.

Choose the correct answer below.

The statement is true. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is true. Every matrix transformation has the property

T(u+v)=T(u)+T(v),

but not all matrix transformations have the property

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is false. Every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

The statement is false. Not every matrix transformation has the properties

T(u+v)=T(u)+T(v)

and

T(cu)=cT(u)

for all u and

v,

in the domain of T and all scalars c.

1.8.23

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×5

matrix and T is a transformation defined by

T(x)=Ax,

then the domain of T is

ℝ3.

Choose the correct answer below.

The statement is false. The domain is actually

ℝ,

because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝ.

The statement is false. The domain is actually

ℝ5,

because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝn.

The statement is true. The domain is

ℝ3

because A has 3 columns, because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝm.

The statement is true. The domain is

ℝ3

because A has 3 rows, because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝm.

1.8.23

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×5

matrix and T is a transformation defined by

T(x)=Ax,

then the domain of T is

ℝ3.

Choose the correct answer below.

The statement is false. The domain is actually

ℝ,

because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝ.

The statement is true. The domain is

ℝ3

because A has 3 columns, because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝm.

The statement is false. The domain is actually

ℝ5,

because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝn.

The statement is true. The domain is

ℝ3

because A has 3 rows, because in the product

Ax,

if A is an

m×n

matrix then x must be a vector in

ℝm.

1.8.25

Determine whether the statement below is true or false. Justify the answer.

If A is an

m×n

matrix, then the range of the transformation

x↦Ax

is

ℝm.

Choose the correct answer below.

The statement is false. The range of the transformation is

ℝn

because the domain of the transformation is

ℝm.

The statement is false. The range of the transformation is the set of all linear combinations of the columns of A, because each image of the transformation is of the form

Ax.

The statement is true. The range of the transformation is

ℝm,

because each vector in

ℝm

is a linear combination of the rows of A.

The statement is true. The range of the transformation is

ℝm,

because each vector in

ℝm

is a linear combination of the columns of A.

1.8.25

Determine whether the statement below is true or false. Justify the answer.

If A is an

m×n

matrix, then the range of the transformation

x↦Ax

is

ℝm.

Choose the correct answer below.

The statement is false. The range of the transformation is the set of all linear combinations of the columns of A, because each image of the transformation is of the form

Ax.

The statement is true. The range of the transformation is

ℝm,

because each vector in

ℝm

is a linear combination of the rows of A.

The statement is false. The range of the transformation is

ℝn

because the domain of the transformation is

ℝm.

The statement is true. The range of the transformation is

ℝm,

because each vector in

ℝm

is a linear combination of the columns of A.

1.8.28

Determine whether the statement below is true or false. Justify the answer.

A linear transformation preserves the operations of vector addition and scalar multiplication.

Choose the correct answer below.

The statement is true. The linear transformation

T(cu+dv)

is not the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are preserved.

The statement is false. The linear transformation

T(cu+dv)

is not the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are not preserved.

The statement is true. The linear transformation

T(cu+dv)

is the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are preserved.

The statement is false. The linear transformation

T(cu+dv)

is the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are not preserved.

1.8.28

Determine whether the statement below is true or false. Justify the answer.

A linear transformation preserves the operations of vector addition and scalar multiplication.

Choose the correct answer below.

The statement is false. The linear transformation

T(cu+dv)

is not the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are not preserved.

The statement is false. The linear transformation

T(cu+dv)

is the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are not preserved.

The statement is true. The linear transformation

T(cu+dv)

is the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are preserved.

The statement is true. The linear transformation

T(cu+dv)

is not the same as

cT(u)+dT(v)

in

ℝm.

Therefore, vector addition and scalar multiplication are preserved.

1.8.29

Determine whether the statement below is true or false. Justify the answer.

A transformation T is linear if and only if

Tc1v1+c2v2=c1Tv1+c2Tv2

for all

v1

and

v2

in the domain of T and for all scalars

c1

and

c2.

Choose the correct answer below.

The statement is true. This equation correctly summarizes the properties necessary for a transformation to be linear.

The statement is false. A transformation T is linear if and only if

T(cu)=cT(u)

for all scalars c and all u in the domain of T.

The statement is false. A transformation T is linear if and only if

T(0)=0.

The statement is false. A transformation T is linear if and only if

T(u+v)=T(u)+T(v)

for all

u,

v in the domain of T.

1.8.29

Determine whether the statement below is true or false. Justify the answer.

A transformation T is linear if and only if

Tc1v1+c2v2=c1Tv1+c2Tv2

for all

v1

and

v2

in the domain of T and for all scalars

c1

and

c2.

Choose the correct answer below.

The statement is true. This equation correctly summarizes the properties necessary for a transformation to be linear.

The statement is false. A transformation T is linear if and only if

T(u+v)=T(u)+T(v)

for all

u,

v in the domain of T.

The statement is false. A transformation T is linear if and only if

T(0)=0.

The statement is false. A transformation T is linear if and only if

T(cu)=cT(u)

for all scalars c and all u in the domain of T.

1.8.29

Determine whether the statement below is true or false. Justify the answer.

A transformation T is linear if and only if

Tc1v1+c2v2=c1Tv1+c2Tv2

for all

v1

and

v2

in the domain of T and for all scalars

c1

and

c2.

Choose the correct answer below.

The statement is true. This equation correctly summarizes the properties necessary for a transformation to be linear.

The statement is false. A transformation T is linear if and only if

T(0)=0.

The statement is false. A transformation T is linear if and only if

T(cu)=cT(u)

for all scalars c and all u in the domain of T.

The statement is false. A transformation T is linear if and only if

T(u+v)=T(u)+T(v)

for all

u,

v in the domain of T.

1.8.31

Let

T : ℝ2→ℝ2

be the linear transformation that reflects each point through the

x1-axis,

such that

A=

	1	0
	0	−1

where

x↦Ax.

Make two sketches that illustrate the two properties of a linear transformation.

Choose the correct graph below that shows the property

T(u+v)=T(u)+T(v).

v

u+v

u

T(u+v)

T(u)

T(v)

x1

x2

v

u+v

u

T(u)

T(v)

T(u+v)

x1

x2

v

u+v

u

T(v)

T(u)

T(u+v)

x1

x2

v

u+v

u

T(v)

T(u)

T(u+v)

x1

x2

Choose the correct graph below that shows the property

T(cu)=cT(u).

u

cu

T(cu)

T(u)

cT(u)

x1

x2

u

cu

T(cu)

T(u)

cT(u)

x1

x2

T(u)

T(cu)

u

cu

cT(u)

x1

x2

u

cu

T(u)

T(cu)

cT(u)

x1

x2

1.8.31

Let

T : ℝ2→ℝ2

be the linear transformation that reflects each point through the

x1-axis,

such that

A=

	1	0
	0	−1

where

x↦Ax.

Make two sketches that illustrate the two properties of a linear transformation.

Choose the correct graph below that shows the property

T(u+v)=T(u)+T(v).

v

u+v

u

T(v)

T(u)

T(u+v)

x1

x2

v

u+v

u

T(u)

T(v)

T(u+v)

x1

x2

v

u+v

u

T(u+v)

T(u)

T(v)

x1

x2

v

u+v

u

T(v)

T(u)

T(u+v)

x1

x2

Choose the correct graph below that shows the property

T(cu)=cT(u).

u

cu

T(u)

T(cu)

cT(u)

x1

x2

u

cu

T(cu)

T(u)

cT(u)

x1

x2

u

cu

T(cu)

T(u)

cT(u)

x1

x2

T(u)

T(cu)

u

cu

cT(u)

x1

x2

1.8.39

Let T:

ℝn→ℝm

be a linear transformation, and let

{v1,

v2,

v3}

be a linearly dependent set in

ℝn.

Explain why the set

{T(v1),

T(v2),

T(v3)}

is linearly dependent.

Choose the correct answer below.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)≠0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)=0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)=0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3≠0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)≠0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

1.8.39

Let T:

ℝn→ℝm

be a linear transformation, and let

{v1,

v2,

v3}

be a linearly dependent set in

ℝn.

Explain why the set

{T(v1),

T(v2),

T(v3)}

is linearly dependent.

Choose the correct answer below.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)≠0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)=0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

not all zero, such that

c1v1+c2v2+c3v3≠0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)≠0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

Given that the set

{v1,

v2,

v3}

is linearly dependent, there exist

c1,

c2,

c3,

all zero, such that

c1v1+c2v2+c3v3=0.

It follows that

c1T(v1)+c2T(v2)+c3T(v3)=0.

Therefore, the set

T(v1),

T(v2),

T(v3)}

is linearly dependent.

1.9.1

Assume that T is a linear transformation. Find the standard matrix of T.

T:

ℝ2→ℝ4,

Te1=(2,

1,

2,

1),

and

Te2=(−9,

5,

0, 0), where

e1=(1,0)

and

e2=(0,1).

A=

	2	−9
	1	5
	2	0
	1	0

(Type an integer or decimal for each matrix element.)

1.9.1

Assume that T is a linear transformation. Find the standard matrix of T.

T:

ℝ2→ℝ4,

Te1=(2,

1,

2,

1),

and

Te2=(−4,

7,

0, 0), where

e1=(1,0)

and

e2=(0,1).

A=

	2	−4
	1	7
	2	0
	1	0

(Type an integer or decimal for each matrix element.)

1.9.1

Assume that T is a linear transformation. Find the standard matrix of T.

T:

ℝ2→ℝ4,

Te1=(4,

1,

4,

1),

and

Te2=(−5,

3,

0, 0), where

e1=(1,0)

and

e2=(0,1).

A=

	4	−5
	1	3
	4	0
	1	0

(Type an integer or decimal for each matrix element.)

1.9.1

Assume that T is a linear transformation. Find the standard matrix of T.

T:

ℝ2→ℝ4,

Te1=(8,

1,

8,

1),

and

Te2=(−7,

5,

0, 0), where

e1=(1,0)

and

e2=(0,1).

A=

	8	−7
	1	5
	8	0
	1	0

(Type an integer or decimal for each matrix element.)

1.9.15

Fill in the missing entries of the matrix, assuming that the equation holds for all values of the variables.

?	?	?
?	?	?
?	?	?

x1

x2

x3

=

3x1−5x2

7x1−4x3

−4x2+5x3

Fill in the missing entries of the matrix below.

3	−5	0
7	0	−4
0	−4	5

x1

x2

x3

=

3x1−5x2

7x1−4x3

−4x2+5x3

1.9.15

Fill in the missing entries of the matrix, assuming that the equation holds for all values of the variables.

?	?	?
?	?	?
?	?	?

x1

x2

x3

=

4x1−3x2

x1−4x3

−7x2+8x3

Fill in the missing entries of the matrix below.

4	−3	0
1	0	−4
0	−7	8

x1

x2

x3

=

4x1−3x2

x1−4x3

−7x2+8x3

1.9.17

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in vectors.

Tx1,x2,x3,x4=x1+4x2, 0, 8x2+x4, x2−x4

A=

	1	4	0	0
	0	0	0	0
	0	8	0	1
	0	1	0	−1

(Type an integer or decimal for each matrix element.)

1.9.17

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in vectors.

Tx1,x2,x3,x4=x1+8x2, 0, 5x2+x4, x2−x4

A=

	1	8	0	0
	0	0	0	0
	0	5	0	1
	0	1	0	−1

(Type an integer or decimal for each matrix element.)

1.9.17

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in vectors.

Tx1,x2,x3,x4=x1+9x2, 0, 6x2+x4, x2−x4

A=

	1	9	0	0
	0	0	0	0
	0	6	0	1
	0	1	0	−1

(Type an integer or decimal for each matrix element.)

1.9.17

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in vectors.

Tx1,x2,x3,x4=x1+5x2, 0, 9x2+x4, x2−x4

A=

	1	5	0	0
	0	0	0	0
	0	9	0	1
	0	1	0	−1

(Type an integer or decimal for each matrix element.)

1.9.19

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in vectors.

Tx1,x2,x3=x1−2x2+7x3, x2−8x3

A=

	1	−2	7
	0	1	−8

(Type an integer or decimal for each matrix element.)

1.9.20

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in a vector.

Tx1,x2,x3,x4=−4x1+x2−x3+x4

                                                                (T:

ℝ4→ℝ)

A=

	−4	1	−1	1

1.9.20

Show that T is a linear transformation by finding a matrix that implements the mapping. Note that

x1,

x2,

... are not vectors but are entries in a vector.

Tx1,x2,x3,x4=3x1+x2−2x3−4x4

                                                               (T:

ℝ4→ℝ)

A=

	3	1	−2	−4

1.9.24

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is one-to-one if each vector in

ℝn

maps onto a unique vector in

ℝm.

Choose the correct answer below.

The statement is true. A mapping T is said to be one-to-one if each x in

ℝn

has at least one image for b in

ℝm.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at least one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each b in

ℝm

is the image of exactly one x in

ℝn.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at most one x in

ℝn.

1.9.24

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is one-to-one if each vector in

ℝn

maps onto a unique vector in

ℝm.

Choose the correct answer below.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at most one x in

ℝn.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at least one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each b in

ℝm

is the image of exactly one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each x in

ℝn

has at least one image for b in

ℝm.

1.9.24

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is one-to-one if each vector in

ℝn

maps onto a unique vector in

ℝm.

Choose the correct answer below.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at least one x in

ℝn.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at most one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each b in

ℝm

is the image of exactly one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each x in

ℝn

has at least one image for b in

ℝm.

1.9.24

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is one-to-one if each vector in

ℝn

maps onto a unique vector in

ℝm.

Choose the correct answer below.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at least one x in

ℝn.

The statement is true. A mapping T is said to be one-to-one if each x in

ℝn

has at least one image for b in

ℝm.

The statement is true. A mapping T is said to be one-to-one if each b in

ℝm

is the image of exactly one x in

ℝn.

The statement is false. A mapping T is said to be one-to-one if each b in

ℝm

is the image of at most one x in

ℝn.

1.9.29

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector x in

ℝn

maps onto some vector in

ℝm.

Choose the correct answer below.

The statement is true. A transformation is onto when all possible values in the codomain, or range, are mapped to by some value in the domain.

The statement is false. A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector in

ℝm

is mapped onto by some vector x in

ℝn.

The statement is false. A mapping T:

ℝn→ℝm

is one-to-one if every vector x in

ℝn

maps onto some vector in

ℝm.

The statement is true. A transformation is onto when, for every b in the codomain, the matrix equation

Ax=b

has a unique solution.

1.9.29

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector x in

ℝn

maps onto some vector in

ℝm.

Choose the correct answer below.

The statement is true. A transformation is onto when all possible values in the codomain, or range, are mapped to by some value in the domain.

The statement is false. A mapping T:

ℝn→ℝm

is one-to-one if every vector x in

ℝn

maps onto some vector in

ℝm.

The statement is false. A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector in

ℝm

is mapped onto by some vector x in

ℝn.

The statement is true. A transformation is onto when, for every b in the codomain, the matrix equation

Ax=b

has a unique solution.

1.9.29

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector x in

ℝn

maps onto some vector in

ℝm.

Choose the correct answer below.

The statement is false. A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector in

ℝm

is mapped onto by some vector x in

ℝn.

The statement is true. A transformation is onto when all possible values in the codomain, or range, are mapped to by some value in the domain.

The statement is false. A mapping T:

ℝn→ℝm

is one-to-one if every vector x in

ℝn

maps onto some vector in

ℝm.

The statement is true. A transformation is onto when, for every b in the codomain, the matrix equation

Ax=b

has a unique solution.

1.9.29

Determine whether the statement below is true or false. Justify the answer.

A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector x in

ℝn

maps onto some vector in

ℝm.

Choose the correct answer below.

The statement is true. A transformation is onto when all possible values in the codomain, or range, are mapped to by some value in the domain.

The statement is false. A mapping T:

ℝn→ℝm

is onto

ℝm

if every vector in

ℝm

is mapped onto by some vector x in

ℝn.

The statement is true. A transformation is onto when, for every b in the codomain, the matrix equation

Ax=b

has a unique solution.

The statement is false. A mapping T:

ℝn→ℝm

is one-to-one if every vector x in

ℝn

maps onto some vector in

ℝm.

1.9.31

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×2

matrix, then the transformation

x↦Ax

cannot be one-to-one.

Choose the correct answer below.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. If

Ax=b

does not have a free variable, then the transformation represented by A is one-to-one.

The statement is true. Transformations which have standard matrices which are not square cannot be one-to-one nor onto because they do not have pivot positions in every row and column.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. It does not matter what dimensions a vector is as long as it meets this requirement. The matrix A could be

1×4

and still represent a one-to-one transformation.

The statement is true. A transformation is one-to-one only if the columns of A are linearly independent and a

3×2

matrix cannot have linearly independent columns.

1.9.31

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×2

matrix, then the transformation

x↦Ax

cannot be one-to-one.

Choose the correct answer below.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. If

Ax=b

does not have a free variable, then the transformation represented by A is one-to-one.

The statement is true. A transformation is one-to-one only if the columns of A are linearly independent and a

3×2

matrix cannot have linearly independent columns.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. It does not matter what dimensions a vector is as long as it meets this requirement. The matrix A could be

1×4

and still represent a one-to-one transformation.

The statement is true. Transformations which have standard matrices which are not square cannot be one-to-one nor onto because they do not have pivot positions in every row and column.

1.9.31

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×2

matrix, then the transformation

x↦Ax

cannot be one-to-one.

Choose the correct answer below.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. It does not matter what dimensions a vector is as long as it meets this requirement. The matrix A could be

1×4

and still represent a one-to-one transformation.

The statement is true. Transformations which have standard matrices which are not square cannot be one-to-one nor onto because they do not have pivot positions in every row and column.

The statement is true. A transformation is one-to-one only if the columns of A are linearly independent and a

3×2

matrix cannot have linearly independent columns.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. If

Ax=b

does not have a free variable, then the transformation represented by A is one-to-one.

1.9.31

Determine whether the statement below is true or false. Justify the answer.

If A is a

3×2

matrix, then the transformation

x↦Ax

cannot be one-to-one.

Choose the correct answer below.

The statement is true. A transformation is one-to-one only if the columns of A are linearly independent and a

3×2

matrix cannot have linearly independent columns.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. If

Ax=b

does not have a free variable, then the transformation represented by A is one-to-one.

The statement is false. A transformation is one-to-one if each vector in the codomain is mapped to by at most one vector in the domain. It does not matter what dimensions a vector is as long as it meets this requirement. The matrix A could be

1×4

and still represent a one-to-one transformation.

The statement is true. Transformations which have standard matrices which are not square cannot be one-to-one nor onto because they do not have pivot positions in every row and column.

1.9.33

Determine if the specified linear transformation is

(a)

one-to-one and

(b)

onto. Justify your answer.

Tx1,x2,x3,x4=(0,x3+x4,x1+x2,x1+x2)

a. Is the linear transformation one-to-one?

T is not one-to-one because the columns of the standard matrix A are linearly independent.

T is one-to-one because

T(x)=0

has only the trivial solution.

T is one-to-one because the column vectors are not scalar multiples of each other.

T is not one-to-one because the standard matrix A has a free variable.

b. Is the linear transformation onto?

T is onto because the standard matrix A does not have a pivot position for every row.

T is onto because the columns of the standard matrix A span

ℝ4.

T is not onto because the

first row

of the standard matrix A is all zeros.

T is not onto because the columns of the standard matrix A span

ℝ4.

1.9.33

Determine if the specified linear transformation is

(a)

one-to-one and

(b)

onto. Justify your answer.

Tx1,x2,x3,x4=(0,x3+x4,x1+x2,x2+x3)

a. Is the linear transformation one-to-one?

T is not one-to-one because the columns of the standard matrix A are linearly independent.

T is one-to-one because the column vectors are not scalar multiples of each other.

T is one-to-one because

T(x)=0

has only the trivial solution.

T is not one-to-one because the standard matrix A has a free variable.

b. Is the linear transformation onto?

T is onto because the standard matrix A does not have a pivot position for every row.

T is not onto because the columns of the standard matrix A span

ℝ4.

T is onto because the columns of the standard matrix A span

ℝ4.

T is not onto because the

first row

of the standard matrix A is all zeros.

1.9.35

Determine if the specified linear transformation is

(a)

one-to-one and

(b)

onto. Justify each answer.

Tx1,x2,x3=x1−5x2+6x3, x2−3x3

(a)

Is the linear transformation one-to-one?

T is not one-to-one because the columns of the standard matrix A are linearly dependent.

T is one-to-one because the column vectors are not scalar multiples of each other.

T is one-to-one because

T(x)=0

has only the trivial solution.

T is not one-to-one because the columns of the standard matrix A are linearly independent.

(b)

Is the linear transformation onto?

T is not onto because the columns of the standard matrix A span

ℝ2.

T is onto because the standard matrix A does not have a pivot position for every row.

T is onto because the columns of the standard matrix A span

ℝ2.

T is not onto because the standard matrix A does not have a pivot position for every row.

1.9.35

Determine if the specified linear transformation is

(a)

one-to-one and

(b)

onto. Justify each answer.

Tx1,x2,x3=x1−7x2+4x3, x2−9x3

(a)

Is the linear transformation one-to-one?

T is one-to-one because the column vectors are not scalar multiples of each other.

T is not one-to-one because the columns of the standard matrix A are linearly dependent.

T is not one-to-one because the columns of the standard matrix A are linearly independent.

T is one-to-one because

T(x)=0

has only the trivial solution.

(b)

Is the linear transformation onto?

T is onto because the columns of the standard matrix A span

ℝ2.

T is onto because the standard matrix A does not have a pivot position for every row.

T is not onto because the columns of the standard matrix A span

ℝ2.

T is not onto because the standard matrix A does not have a pivot position for every row.

1.9.45

Let T be the linear transformation whose standard matrix is given. Decide if T is a one-to-one mapping. Justify your answer.

−7 5 −7 −2 8 5 −4 25 6 10 7 15 −3 1 9 −14

Choose the correct answer below.

The transformation T is not one-to-one because the equation

T(x)=0

has only the trivial solution.

The transformation T is one-to-one because the equation

T(x)=0

has a nontrivial solution.

The transformation T is one-to-one because the equation

T(x)=0

has only the trivial solution.

The transformation T is not one-to-one because the equation

T(x)=0

has a nontrivial solution.

1.9.45

Let T be the linear transformation whose standard matrix is given. Decide if T is a one-to-one mapping. Justify your answer.

−4 6 −4 12 9 2 −5 18 6 12 4 26 −5 1 9 −12

Choose the correct answer below.

The transformation T is not one-to-one because the equation

T(x)=0

has a nontrivial solution.

The transformation T is not one-to-one because the equation

T(x)=0

has only the trivial solution.

The transformation T is one-to-one because the equation

T(x)=0

has a nontrivial solution.

The transformation T is one-to-one because the equation

T(x)=0

has only the trivial solution.

1.9.45

Let T be the linear transformation whose standard matrix is given. Decide if T is a one-to-one mapping. Justify your answer.

−5 6 −5 −3 8 4 −5 6 7 8 5 33 −2 2 11 24

Choose the correct answer below.

The transformation T is one-to-one because the equation

T(x)=0

has only the trivial solution.

The transformation T is not one-to-one because the equation

T(x)=0

has a nontrivial solution.

The transformation T is one-to-one because the equation

T(x)=0

has a nontrivial solution.

The transformation T is not one-to-one because the equation

T(x)=0

has only the trivial solution.

1.10.7

Write a matrix equation that determines the loop currents.

I1

I2

I3

I4

3 Ω

3 Ω

9 Ω

2 Ω

6 Ω

6 Ω

4 Ω

3 Ω

For each matrix, let row 1 correspond to loop 1, row 2 correspond to loop 2, and so on. Also, enter positive values for positive voltages and negative values for negative voltages.

	15	−9	0	−3
	−9	17	−6	0
	0	−6	15	−6
	−3	0	−6	13

	I1
	I2
	I3
	I4

=

	41
	25
	24
	−11

1.10.7

Write a matrix equation that determines the loop currents.

I1

I2

I3

I4

1 Ω

3 Ω

8 Ω

2 Ω

4 Ω

6 Ω

5 Ω

4 Ω

For each matrix, let row 1 correspond to loop 1, row 2 correspond to loop 2, and so on. Also, enter positive values for positive voltages and negative values for negative voltages.

	12	−8	0	−3
	−8	16	−6	0
	0	−6	14	−4
	−3	0	−4	12

	I1
	I2
	I3
	I4

=

	39
	28
	18
	−12

1.10.8

Write a matrix equation that determines the loop currents shown to the right.

I1 I2 I4 I3 I5

Complete the matrix equation.

	12	−2	0	−2	−4
	−2	13	−4	0	−4
	0	−4	17	−5	−5
	−2	0	−5	11	−3
	−4	−4	−5	−3	16

	I1
	I2
	I3
	I4
	I5

=

	51
	−38
	24
	−48
	0

1.10.8

Write a matrix equation that determines the loop currents shown to the right.

I1 I2 I4 I3 I5

Complete the matrix equation.

	12	−2	0	−2	−4
	−2	11	−4	0	−4
	0	−4	13	−3	−3
	−2	0	−3	12	−4
	−4	−4	−3	−4	15

	I1
	I2
	I3
	I4
	I5

=

	51
	−35
	29
	−40
	0

1.10.8

Write a matrix equation that determines the loop currents shown to the right.

I1 I2 I4 I3 I5

Complete the matrix equation.

	8	−1	0	−2	−3
	−1	9	−4	0	−2
	0	−4	13	−3	−4
	−2	0	−3	9	−2
	−3	−2	−4	−2	11

	I1
	I2
	I3
	I4
	I5

=

	52
	−38
	24
	−45
	0

1.10.8

Write a matrix equation that determines the loop currents shown to the right.

I1 I2 I4 I3 I5

Complete the matrix equation.

	9	−1	0	−1	−3
	−1	9	−3	0	−3
	0	−3	14	−4	−5
	−1	0	−4	9	−3
	−3	−3	−5	−3	14

	I1
	I2
	I3
	I4
	I5

=

	55
	−33
	28
	−40
	0

1.10.12

A car rental service in a certain town has a fleet of about

700

cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix to the right. Suppose that on Monday there are

355

cars at the airport (or rented from there),

75

cars at the east side office, and

270

cars at the west side office. What will be the approximate distribution of cars on Wednesday?

Cars Rented From:
Airport	East	West	Returned To:
0.95       0.07       0.12 0.00       0.86       0.06 0.05       0.07       0.82			Airport
			East
			West

On Wednesday,

391

cars will be at the airport,

84

cars will be at the East location, and

225

cars will be at the West location.

(Round to the nearest integer as needed.)

1.10.12

A car rental service in a certain town has a fleet of about

600

cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix to the right. Suppose that on Monday there are

195

cars at the airport (or rented from there),

55

cars at the east side office, and

350

cars at the west side office. What will be the approximate distribution of cars on Wednesday?

Cars Rented From:
Airport	East	West	Returned To:
0.95       0.04       0.10 0.00       0.92       0.05 0.05       0.04       0.85			Airport
			East
			West

On Wednesday,

245

cars will be at the airport,

78

cars will be at the East location, and

277

cars will be at the West location.

(Round to the nearest integer as needed.)

2.1.1

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	2	0	−1
	4	−6	4

,

B=

	7	−4	2
	1	−5	−2

,

C=

	2	3
	−2	2

,

and

D=

	4	6
	−2	5

.

−3A,

B−3A,

AC, CD

Compute the matrix product

−3A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

−3A=

	−6	0	3
	−12	18	−12

(Simplify your answer.)

The expression

−3A

is undefined because A is not a square matrix.

The expression

−3A

is undefined because matrices cannot be multiplied by numbers.

The expression

−3A

is undefined because matrices cannot have negative coefficients.

Compute the martrix sum

B−3A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

B−3A=

	1	−4	5
	−11	13	−14

(Simplify your answer.)

The expression

B−3A

is undefined because B and A have different sizes.

The expression

B−3A

is undefined because B and

−3A

have different sizes.

The expression

B−3A

is undefined because A is not a square matrix.

Compute the matrix product AC. Select the correct choice below and, if necessary, fill in the answer box within your choice.

AC=

(Simplify your answer.)

The expression AC is undefined because the number of rows in A is not equal to the number of rows in C.

The expression AC is undefined because the number of columns in A is not equal to the number of rows in C.

The expression AC is undefined because the number of rows in A is not equal to the number of columns in C.

Compute the matrix product CD. Select the correct choice below and, if necessary, fill in the answer box within your choice.

CD=

	2	27
	−12	−2

(Simplify your answer.)

The expression CD is undefined because the corresponding entries in C and D are not equal.

The expression CD is undefined because matrices with negative entries cannot be multiplied.

The expression CD is undefined because square matrices cannot be multiplied.

2.1.1

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	3	0	−1
	3	−6	3

,

B=

	6	−6	1
	2	−5	−3

,

C=

	1	3
	−1	1

,

and

D=

	4	6
	−2	4

.

−2A,

B−2A,

AC, CD

Compute the matrix product

−2A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

−2A=

	−6	0	2
	−6	12	−6

(Simplify your answer.)

The expression

−2A

is undefined because matrices cannot have negative coefficients.

The expression

−2A

is undefined because A is not a square matrix.

The expression

−2A

is undefined because matrices cannot be multiplied by numbers.

Compute the martrix sum

B−2A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

B−2A=

	0	−6	3
	−4	7	−9

(Simplify your answer.)

The expression

B−2A

is undefined because A is not a square matrix.

The expression

B−2A

is undefined because B and A have different sizes.

The expression

B−2A

is undefined because B and

−2A

have different sizes.

Compute the matrix product AC. Select the correct choice below and, if necessary, fill in the answer box within your choice.

AC=

(Simplify your answer.)

The expression AC is undefined because the number of rows in A is not equal to the number of columns in C.

The expression AC is undefined because the number of columns in A is not equal to the number of rows in C.

The expression AC is undefined because the number of rows in A is not equal to the number of rows in C.

Compute the matrix product CD. Select the correct choice below and, if necessary, fill in the answer box within your choice.

CD=

	−2	18
	−6	−2

(Simplify your answer.)

The expression CD is undefined because square matrices cannot be multiplied.

The expression CD is undefined because matrices with negative entries cannot be multiplied.

The expression CD is undefined because the corresponding entries in C and D are not equal.

2.1.1

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	1	0	−1
	4	−6	3

,

B=

	8	−4	2
	2	−4	−2

,

C=

	2	3
	−1	2

,

and

D=

	4	5
	−1	4

.

−3A,

B−3A,

AC, CD

Compute the matrix product

−3A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

−3A=

	−3	0	3
	−12	18	−9

(Simplify your answer.)

The expression

−3A

is undefined because matrices cannot have negative coefficients.

The expression

−3A

is undefined because A is not a square matrix.

The expression

−3A

is undefined because matrices cannot be multiplied by numbers.

Compute the martrix sum

B−3A.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

B−3A=

	5	−4	5
	−10	14	−11

(Simplify your answer.)

The expression

B−3A

is undefined because B and A have different sizes.

The expression

B−3A

is undefined because B and

−3A

have different sizes.

The expression

B−3A

is undefined because A is not a square matrix.

Compute the matrix product AC. Select the correct choice below and, if necessary, fill in the answer box within your choice.

AC=

(Simplify your answer.)

The expression AC is undefined because the number of rows in A is not equal to the number of rows in C.

The expression AC is undefined because the number of columns in A is not equal to the number of rows in C.

The expression AC is undefined because the number of rows in A is not equal to the number of columns in C.

Compute the matrix product CD. Select the correct choice below and, if necessary, fill in the answer box within your choice.

CD=

	5	22
	−6	3

(Simplify your answer.)

The expression CD is undefined because matrices with negative entries cannot be multiplied.

The expression CD is undefined because the corresponding entries in C and D are not equal.

The expression CD is undefined because square matrices cannot be multiplied.

2.1.2

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	1	0	−3
	3	−6	4

,

B=

	7	−4	2
	2	−3	−3

,

C=

	1	3
	−2	1

,

D=

	3	4
	−2	5

,

and

E=

	−5
	2

.

A+3B,

4C−3E,

DB, EB

Compute the matrix sum

A+3B.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

A+3B=

	22	−12	3
	9	−15	−5

(Simplify your answer.)

The expression

A+3B

is undefined because B is not a square matrix.

The expression

A+3B

is undefined because A and

3B

have different sizes.

The expression

A+3B

is undefined because A is not a square matrix.

Compute the matrix sum

4C−3E.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

4C−3E=

(Simplify your answer.)

The expression

4C−3E

is undefined because

4C

and

3E

have different sizes.

The expression

4C−3E

is undefined because E is not a square matrix.

The expression

4C−3E

is undefined because the number of rows in C is not equal to the number of columns in E.

Compute the matrix product DB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

DB=

	29	−24	−6
	−4	−7	−19

(Simplify your answer.)

The expression DB is undefined because the number of rows in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of rows in B.

Compute the matrix product EB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

EB=

(Simplify your answer.)

The expression EB is undefined because the number of columns in E is not equal to the number of rows in B.

The expression EB is undefined because the number of columns in E is not equal to the number of columns in B.

The expression EB is undefined because the number of rows in E is not equal to the number of columns in B.

2.1.2

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	1	0	−1
	5	−4	2

,

B=

	8	−4	1
	2	−4	−3

,

C=

	1	3
	−2	1

,

D=

	4	6
	−2	4

,

and

E=

	−5
	4

.

A+4B,

2C−4E,

DB, EB

Compute the matrix sum

A+4B.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

A+4B=

	33	−16	3
	13	−20	−10

(Simplify your answer.)

The expression

A+4B

is undefined because B is not a square matrix.

The expression

A+4B

is undefined because A and

4B

have different sizes.

The expression

A+4B

is undefined because A is not a square matrix.

Compute the matrix sum

2C−4E.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

2C−4E=

(Simplify your answer.)

The expression

2C−4E

is undefined because the number of rows in C is not equal to the number of columns in E.

The expression

2C−4E

is undefined because E is not a square matrix.

The expression

2C−4E

is undefined because

2C

and

4E

have different sizes.

Compute the matrix product DB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

DB=

	44	−40	−14
	−8	−8	−14

(Simplify your answer.)

The expression DB is undefined because the number of rows in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of rows in B.

Compute the matrix product EB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

EB=

(Simplify your answer.)

The expression EB is undefined because the number of rows in E is not equal to the number of columns in B.

The expression EB is undefined because the number of columns in E is not equal to the number of rows in B.

The expression EB is undefined because the number of columns in E is not equal to the number of columns in B.

2.1.2

Compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

A=

	2	0	−1
	4	−5	3

,

B=

	6	−4	2
	2	−3	−4

,

C=

	2	3
	−1	2

,

D=

	2	6
	−2	4

,

and

E=

	−5
	4

.

A+3B,

3C−2E,

DB, EB

Compute the matrix sum

A+3B.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

A+3B=

	20	−12	5
	10	−14	−9

(Simplify your answer.)

The expression

A+3B

is undefined because B is not a square matrix.

The expression

A+3B

is undefined because A is not a square matrix.

The expression

A+3B

is undefined because A and

3B

have different sizes.

Compute the matrix sum

3C−2E.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

3C−2E=

(Simplify your answer.)

The expression

3C−2E

is undefined because the number of rows in C is not equal to the number of columns in E.

The expression

3C−2E

is undefined because E is not a square matrix.

The expression

3C−2E

is undefined because

3C

and

2E

have different sizes.

Compute the matrix product DB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

DB=

	24	−26	−20
	−4	−4	−20

(Simplify your answer.)

The expression DB is undefined because the number of rows in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of columns in B.

The expression DB is undefined because the number of columns in D is not equal to the number of rows in B.

Compute the matrix product EB. Select the correct choice below and, if necessary, fill in the answer box within your choice.

EB=

(Simplify your answer.)

The expression EB is undefined because the number of columns in E is not equal to the number of columns in B.

The expression EB is undefined because the number of columns in E is not equal to the number of rows in B.

The expression EB is undefined because the number of rows in E is not equal to the number of columns in B.

2.1.4

Compute

A−3I3

and

(3I3)A,

where

A=

	6	−2	3
	−3	4	−6
	−3	1	3

.

A−3I3=

	3	−2	3
	−3	1	−6
	−3	1	0

(3I3)A=

	18	−6	9
	−9	12	−18
	−9	3	9

2.1.4

Compute

A−3I3

and

(3I3)A,

where

A=

	6	−1	2
	−4	3	−5
	−2	1	1

.

A−3I3=

	3	−1	2
	−4	0	−5
	−2	1	−2

(3I3)A=

	18	−3	6
	−12	9	−15
	−6	3	3

2.1.4

Compute

A−5I3

and

(5I3)A,

where

A=

	6	−2	2
	−5	2	−7
	−4	2	2

.

A−5I3=

	1	−2	2
	−5	−3	−7
	−4	2	−3

(5I3)A=

	30	−10	10
	−25	10	−35
	−20	10	10

2.1.5

Compute the product AB by the definition of the product of matrices, where

Ab1

and

Ab2

are computed separately, and by the row-column rule for computing AB.

A=

	−2	3
	1	4
	5	−4

,

B=

	5	−3
	−1	3

Set up the product

Ab1,

where

b1

is the first column of B.

Ab1=

	−2	3
	1	4
	5	−4

	5
	−1

(Use one answer box for A and use the other answer box for

b1.)

Calculate

Ab1,

where

b1

is the first column of B.

Ab1=

	−13
	1
	29

(Type an integer or decimal for each matrix element.)

Set up the product

Ab2,

where

b2

is the second column of B.

Ab2=

	−2	3
	1	4
	5	−4

	−3
	3

(Use one answer box for A and use the other answer box for

b2.)

Calculate

Ab2,

where

b2

is the second column of B.

Ab2=

	15
	9
	−27

(Type an integer or decimal for each matrix element.)

Determine the numerical expression for the first entry in the first column of AB using the row-column rule. Choose the correct answer below.

((−2)−(5))•((3)−(−1)

−2(5)−3(−1)

((−2)+(5))•((3)+(−1)

−2(5)+3(−1)

Determine the product AB.

AB=

	−13	15
	1	9
	29	−27

(Use integers or decimals for any numbers in the expression.)

2.1.5

Compute the product AB by the definition of the product of matrices, where

Ab1

and

Ab2

are computed separately, and by the row-column rule for computing AB.

A=

	−1	3
	3	3
	4	−3

,

B=

	5	−1
	−1	2

Set up the product

Ab1,

where

b1

is the first column of B.

Ab1=

	−1	3
	3	3
	4	−3

	5
	−1

(Use one answer box for A and use the other answer box for

b1.)

Calculate

Ab1,

where

b1

is the first column of B.

Ab1=

	−8
	12
	23

(Type an integer or decimal for each matrix element.)

Set up the product

Ab2,

where

b2

is the second column of B.

Ab2=

	−1	3
	3	3
	4	−3

	−1
	2

(Use one answer box for A and use the other answer box for

b2.)

Calculate

Ab2,

where

b2

is the second column of B.

Ab2=

	7
	3
	−10

(Type an integer or decimal for each matrix element.)

Determine the numerical expression for the first entry in the first column of AB using the row-column rule. Choose the correct answer below.

−1(5)−3(−1)

((−1)+(5))•((3)+(−1)

((−1)−(5))•((3)−(−1)

−1(5)+3(−1)

Determine the product AB.

AB=

	−8	7
	12	3
	23	−10

(Use integers or decimals for any numbers in the expression.)

2.1.5

Compute the product AB by the definition of the product of matrices, where

Ab1

and

Ab2

are computed separately, and by the row-column rule for computing AB.

A=

	−2	2
	1	5
	6	−3

,

B=

	4	−3
	−3	3

Set up the product

Ab1,

where

b1

is the first column of B.

Ab1=

	−2	2
	1	5
	6	−3

	4
	−3

(Use one answer box for A and use the other answer box for

b1.)

Calculate

Ab1,

where

b1

is the first column of B.

Ab1=

	−14
	−11
	33

(Type an integer or decimal for each matrix element.)

Set up the product

Ab2,

where

b2

is the second column of B.

Ab2=

	−2	2
	1	5
	6	−3

	−3
	3

(Use one answer box for A and use the other answer box for

b2.)

Calculate

Ab2,

where

b2

is the second column of B.

Ab2=

	12
	12
	−27

(Type an integer or decimal for each matrix element.)

Determine the numerical expression for the first entry in the first column of AB using the row-column rule. Choose the correct answer below.

−2(4)−2(−3)

−2(4)+2(−3)

((−2)−(4))•((2)−(−3)

((−2)+(4))•((2)+(−3)

Determine the product AB.

AB=

	−14	12
	−11	12
	33	−27

(Use integers or decimals for any numbers in the expression.)

2.1.7

If a matrix A is

5×2

and the product AB is

5×7,

what is the size of B?

The size of B is

2×7.

2.1.7

If a matrix A is

3×2

and the product AB is

3×7,

what is the size of B?

The size of B is

2×7.

2.1.8

How many rows does B have if BC is a

4×3

matrix?

Matrix B has

4

rows.

2.1.8

How many rows does B have if BC is a

9×7

matrix?

Matrix B has

9

rows.

2.1.9

Let

A=

2	3
−1	1

and

B=

2	9
−3	k

.

What value(s) of k, if any, will make

AB=BA?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

k=−1

(Use a comma to separate answers as needed.)

No value of k will make

AB=BA

2.1.9

Let

A=

3	4
−1	1

and

B=

2	8
−2	k

.

What value(s) of k, if any, will make

AB=BA?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

k=−2

(Use a comma to separate answers as needed.)

No value of k will make

AB=BA

2.1.9

Let

A=

3	2
−1	2

and

B=

2	6
−3	k

.

What value(s) of k, if any, will make

AB=BA?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

k=−1

(Use a comma to separate answers as needed.)

No value of k will make

AB=BA

2.1.9

Let

A=

4	2
−1	2

and

B=

2	4
−2	k

.

What value(s) of k, if any, will make

AB=BA?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

k=−2

(Use a comma to separate answers as needed.)

No value of k will make

AB=BA

2.1.10

Let

A=

	−4	6
	8	−12

,

B=

	6	2
	4	7

,

and

C=

	3	−1
	2	5

.

Verify that

AB=AC

and yet

B≠C.

Show the calculations that are used to find the entries for matrix AB. Choose the correct answer below.

	−4(6)+(−12)(7)	−4(2)+(−12)(4)
	8(6)+6(7)	8(2)+6(4)

	8(6)+6(4)	8(2)+6(7)
	−4(6)+(−12)(4)	−4(2)+(−12)(7)

	−4(6)+6(4)	−4(2)+6(7)
	8(6)+(−12)(4)	8(2)+(−12)(7)

	−4(2)+6(4)	−4(6)+6(7)
	8(2)+(−12)(4)	8(6)+(−12)(7)

Show the calculations that are used to find the entries for matrix AC. Choose the correct answer below.

	8(3)+6(2)	8(−1)+6(5)
	−4(3)+(−12)(2)	−4(−1)+(−12)(5)

	−4(3)+6(2)	−4(−1)+6(5)
	8(3)+(−12)(2)	8(−1)+(−12)(5)

	−4(−1)+6(2)	−4(3)+6(5)
	8(−1)+(−12)(2)	8(3)+(−12)(5)

	−4(3)+(−12)(5)	−4(−1)+(−12)(2)
	8(3)+6(5)	8(−1)+6(2)

Verify that

AB=AC

by simplifying.

AB=AC=

	0	34
	0	−68

(Type an integer or decimal for each matrix element.)

2.1.10

Let

A=

	4	−6
	−8	12

,

B=

	8	4
	6	9

,

and

C=

	2	−2
	2	5

.

Verify that

AB=AC

and yet

B≠C.

Show the calculations that are used to find the entries for matrix AB. Choose the correct answer below.

	4(4)+(−6)(6)	4(8)+(−6)(9)
	−8(4)+12(6)	−8(8)+12(9)

	−8(8)+(−6)(6)	−8(4)+(−6)(9)
	4(8)+12(6)	4(4)+12(9)

	4(8)+12(9)	4(4)+12(6)
	−8(8)+(−6)(9)	−8(4)+(−6)(6)

	4(8)+(−6)(6)	4(4)+(−6)(9)
	−8(8)+12(6)	−8(4)+12(9)

Show the calculations that are used to find the entries for matrix AC. Choose the correct answer below.

	4(−2)+(−6)(2)	4(2)+(−6)(5)
	−8(−2)+12(2)	−8(2)+12(5)

	4(2)+12(5)	4(−2)+12(2)
	−8(2)+(−6)(5)	−8(−2)+(−6)(2)

	−8(2)+(−6)(2)	−8(−2)+(−6)(5)
	4(2)+12(2)	4(−2)+12(5)

	4(2)+(−6)(2)	4(−2)+(−6)(5)
	−8(2)+12(2)	−8(−2)+12(5)

Verify that

AB=AC

by simplifying.

AB=AC=

	−4	−38
	8	76

(Type an integer or decimal for each matrix element.)

2.1.10

Let

A=

	−4	6
	8	−12

,

B=

	5	4
	7	9

,

and

C=

	−1	−2
	3	5

.

Verify that

AB=AC

and yet

B≠C.

Show the calculations that are used to find the entries for matrix AB. Choose the correct answer below.

	−4(5)+6(7)	−4(4)+6(9)
	8(5)+(−12)(7)	8(4)+(−12)(9)

	−4(5)+(−12)(9)	−4(4)+(−12)(7)
	8(5)+6(9)	8(4)+6(7)

	8(5)+6(7)	8(4)+6(9)
	−4(5)+(−12)(7)	−4(4)+(−12)(9)

	−4(4)+6(7)	−4(5)+6(9)
	8(4)+(−12)(7)	8(5)+(−12)(9)

Show the calculations that are used to find the entries for matrix AC. Choose the correct answer below.

	−4(−1)+6(3)	−4(−2)+6(5)
	8(−1)+(−12)(3)	8(−2)+(−12)(5)

	−4(−2)+6(3)	−4(−1)+6(5)
	8(−2)+(−12)(3)	8(−1)+(−12)(5)

	8(−1)+6(3)	8(−2)+6(5)
	−4(−1)+(−12)(3)	−4(−2)+(−12)(5)

	−4(−1)+(−12)(5)	−4(−2)+(−12)(3)
	8(−1)+6(5)	8(−2)+6(3)

Verify that

AB=AC

by simplifying.

AB=AC=

	22	38
	−44	−76

(Type an integer or decimal for each matrix element.)

2.1.11

Let

A=

1	1	1
1	4	5
1	5	6

and

D=

6	0	0
0	5	0
0	0	2

.

Compute AD and DA. Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Compute AD.

AD=

	6	5	2
	6	20	10
	6	25	12

Compute DA.

DA=

	6	6	6
	5	20	25
	2	10	12

Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Choose the correct answer below.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each column entry of A by the corresponding diagonal entry of D.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each row of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each column of A by the corresponding diagonal entry of D.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each row entry of A by the corresponding diagonal entry of D.

Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Choose the correct answer below.

There is only one unique solution,

B=.

(Simplify your answers.)

There are infinitely many solutions. Any multiple of

I3

will satisfy the expression.

There does not exist a matrix, B, that will satisfy the expression.

2.1.11

Let

A=

1	1	1
1	5	7
1	7	6

and

D=

5	0	0
0	3	0
0	0	2

.

Compute AD and DA. Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Compute AD.

AD=

	5	3	2
	5	15	14
	5	21	12

Compute DA.

DA=

	5	5	5
	3	15	21
	2	14	12

Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Choose the correct answer below.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each row entry of A by the corresponding diagonal entry of D.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each column entry of A by the corresponding diagonal entry of D.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each row of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each column of A by the corresponding diagonal entry of D.

Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Choose the correct answer below.

There is only one unique solution,

B=.

(Simplify your answers.)

There are infinitely many solutions. Any multiple of

I3

will satisfy the expression.

There does not exist a matrix, B, that will satisfy the expression.

2.1.11

Let

A=

1	1	1
1	4	6
1	6	7

and

D=

7	0	0
0	4	0
0	0	3

.

Compute AD and DA. Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Compute AD.

AD=

	7	4	3
	7	16	18
	7	24	21

Compute DA.

DA=

	7	7	7
	4	16	24
	3	18	21

Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Choose the correct answer below.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each row of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each column of A by the corresponding diagonal entry of D.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each row entry of A by the corresponding diagonal entry of D.

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D.

Both right-multiplication (that is, multiplication on the right) and left-multiplication by the diagonal matrix D multiplies each column entry of A by the corresponding diagonal entry of D.

Find a

3×3

matrix B, not the identity matrix or zero matrix, such that

AB=BA.

Choose the correct answer below.

There is only one unique solution,

B=.

(Simplify your answers.)

There are infinitely many solutions. Any multiple of

I3

will satisfy the expression.

There does not exist a matrix, B, that will satisfy the expression.

2.1.12

Let

A=

	3	−6
	−5	10

.

Construct a

2×2

matrix B such that AB is the zero matrix. Use two different nonzero columns for B.

B=

	2	4
	1	2

2.1.12

Let

A=

	5	−15
	−3	9

.

Construct a

2×2

matrix B such that AB is the zero matrix. Use two different nonzero columns for B.

B=

	6	3
	2	1

2.1.16

Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer.

If A and B are

3×3

matrices and

B=

b1

b2

b3

,

then

AB=

Ab1+Ab2+Ab3

.

Choose the correct answer below.

The statement is false. The matrix

Ab1+Ab2+Ab3

is the correct size matrix, but the plus signs should be minus signs.

The statement is false. The matrix

Ab1+Ab2+Ab3

is a

3×1

matrix, and AB must be a

3×3

matrix. The plus signs should be spaces between the 3 columns.

The statement is true. By the definition of matrix multiplication, if A is an

m×n

matrix and B is an

n×p

matrix, then the resulting matrix AB is the sum of the columns of A using the weights from the corresponding columns of B.

The statement is true. By the definition of matrix multiplication, if A is an

m×n

matrix and B is an

n×p

matrix, then

AB=A

b1

b2

...

bp

=

Ab1+Ab2+...+Abp

.

2.1.16

Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer.

If A and B are

3×3

matrices and

B=

b1

b2

b3

,

then

AB=

Ab1+Ab2+Ab3

.

Choose the correct answer below.

The statement is false. The matrix

Ab1+Ab2+Ab3

is the correct size matrix, but the plus signs should be minus signs.

The statement is true. By the definition of matrix multiplication, if A is an

m×n

matrix and B is an

n×p

matrix, then

AB=A

b1

b2

...

bp

=

Ab1+Ab2+...+Abp

.

The statement is true. By the definition of matrix multiplication, if A is an

m×n

matrix and B is an

n×p

matrix, then the resulting matrix AB is the sum of the columns of A using the weights from the corresponding columns of B.

The statement is false. The matrix

Ab1+Ab2+Ab3

is a

3×1

matrix, and AB must be a

3×3

matrix. The plus signs should be spaces between the 3 columns.

2.1.19

Let A, B, and C be arbitrary matrices for which the indicated sums and products are defined. Determine whether the statement below is true or false. Justify the answer.

AB+AC=A(B+C)

Choose the correct answer below.

The statement is false. The distributive law does not apply to matrix multiplication.

The statement is true. The multiplication laws for matrices are the same as those for real numbers.

The statement is false. The distributive law for matrices states that

A(B+C)=BA+CA.

The statement is true. The distributive law for matrices states that

A(B+C)=AB+AC.

2.1.19

Let A, B, and C be arbitrary matrices for which the indicated sums and products are defined. Determine whether the statement below is true or false. Justify the answer.

AB+AC=A(B+C)

Choose the correct answer below.

The statement is true. The multiplication laws for matrices are the same as those for real numbers.

The statement is true. The distributive law for matrices states that

A(B+C)=AB+AC.

The statement is false. The distributive law does not apply to matrix multiplication.

The statement is false. The distributive law for matrices states that

A(B+C)=BA+CA.

2.1.20

Let A and B be arbitrary matrices for which the indicated sum is defined. Determine whether the statement below is true or false. Justify the answer.

AT+BT=(A+B)T

Choose the correct answer below.

The statement is false. The transpose property states that

(A+B)T=ATBT.

The statement is true. The transpose property for matrices is the same as for algebraic exponents of real numbers.

The statement is true. The transpose property states that

(A+B)T=AT+BT.

The statement is false. The transpose property is inapplicable here.

2.1.20

Let A and B be arbitrary matrices for which the indicated sum is defined. Determine whether the statement below is true or false. Justify the answer.

AT+BT=(A+B)T

Choose the correct answer below.

The statement is true. The transpose property for matrices is the same as for algebraic exponents of real numbers.

The statement is false. The transpose property states that

(A+B)T=ATBT.

The statement is false. The transpose property is inapplicable here.

The statement is true. The transpose property states that

(A+B)T=AT+BT.

2.1.20

Let A and B be arbitrary matrices for which the indicated sum is defined. Determine whether the statement below is true or false. Justify the answer.

AT+BT=(A+B)T

Choose the correct answer below.

The statement is false. The transpose property is inapplicable here.

The statement is true. The transpose property states that

(A+B)T=AT+BT.

The statement is false. The transpose property states that

(A+B)T=ATBT.

The statement is true. The transpose property for matrices is the same as for algebraic exponents of real numbers.

2.1.21

Let A, B, and C be arbitrary matrices for which the indicated products are defined. Determine whether the statement below is true or false. Justify the answer.

(AB)C=(AC)B

Choose the correct answer below.

The statement is false. The associative law of multiplication for matrices states that

(AB)C=B(AC).

The statement is false. The associative law of multiplication for matrices states that

A(BC)=(AB)C.

The statement is true. The associative law of multiplication for matrices states that

(AB)C=(AC)B.

The statement is true. The associative law of multiplication for matrices states that

(AB)C=B(AC).

2.1.21

Let A, B, and C be arbitrary matrices for which the indicated products are defined. Determine whether the statement below is true or false. Justify the answer.

(AB)C=(AC)B

Choose the correct answer below.

The statement is false. The associative law of multiplication for matrices states that

A(BC)=(AB)C.

The statement is false. The associative law of multiplication for matrices states that

(AB)C=B(AC).

The statement is true. The associative law of multiplication for matrices states that

(AB)C=(AC)B.

The statement is true. The associative law of multiplication for matrices states that

(AB)C=B(AC).

2.1.21

Let A, B, and C be arbitrary matrices for which the indicated products are defined. Determine whether the statement below is true or false. Justify the answer.

(AB)C=(AC)B

Choose the correct answer below.

The statement is true. The associative law of multiplication for matrices states that

(AB)C=(AC)B.

The statement is false. The associative law of multiplication for matrices states that

(AB)C=B(AC).

The statement is true. The associative law of multiplication for matrices states that

(AB)C=B(AC).

The statement is false. The associative law of multiplication for matrices states that

A(BC)=(AB)C.

2.1.22

Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer.

(AB)T=ATBT

Choose the correct answer below.

The statement is false. The transpose of the product of two matrices is the product of the transposes of the individual matrices in reverse order, or

(AB)T=BTAT.

The statement is true. The transpose of the product of two matrices is the product of the transposes of the individual matrices in the same order, or

(AB)T=ATBT.

The statement is true. Matrix multiplication is not commutative so the products must remain in the same order.

The statement is false. The transpose of the product of two matrices is the product of the transpose of the first matrix and the second matrix, or

(AB)T=ATB.

2.1.22

Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer.

(AB)T=ATBT

Choose the correct answer below.

The statement is true. Matrix multiplication is not commutative so the products must remain in the same order.

The statement is false. The transpose of the product of two matrices is the product of the transposes of the individual matrices in reverse order, or

(AB)T=BTAT.

The statement is true. The transpose of the product of two matrices is the product of the transposes of the individual matrices in the same order, or

(AB)T=ATBT.

The statement is false. The transpose of the product of two matrices is the product of the transpose of the first matrix and the second matrix, or

(AB)T=ATB.

2.1.22

Let A and B be arbitrary matrices for which the indicated product is defined. Determine whether the statement below is true or false. Justify the answer.

(AB)T=ATBT

Choose the correct answer below.

The statement is false. The transpose of the product of two matrices is the product of the transposes of the individual matrices in reverse order, or

(AB)T=BTAT.

The statement is true. The transpose of the product of two matrices is the product of the transposes of the individual matrices in the same order, or

(AB)T=ATBT.

The statement is true. Matrix multiplication is not commutative so the products must remain in the same order.

The statement is false. The transpose of the product of two matrices is the product of the transpose of the first matrix and the second matrix, or

(AB)T=ATB.

2.1.26

Suppose the first two columns,

b1

and

b2,

of B are equal. What can you say about the columns of AB (if AB is defined)? Why?

Choose the correct answer below.

The first two columns of AB will be equal only if the first two rows of A are equal.

The first two columns of AB are

Ab1

and

Ab2.

They are equal since

b1

and

b2

are equal.

The first two columns of AB will be equal only if the first two columns of A are equal.

Nothing can be determined about the columns of AB since the entries of A are unknown.

2.1.26

Suppose the first two columns,

b1

and

b2,

of B are equal. What can you say about the columns of AB (if AB is defined)? Why?

Choose the correct answer below.

The first two columns of AB are

Ab1

and

Ab2.

They are equal since

b1

and

b2

are equal.

The first two columns of AB will be equal only if the first two columns of A are equal.

Nothing can be determined about the columns of AB since the entries of A are unknown.

The first two columns of AB will be equal only if the first two rows of A are equal.

2.1.47

Let

S=

	0	9	0	0	0
	0	0	9	0	0
	0	0	0	9	0
	0	0	0	0	9
	0	0	0	0	0

.

Compute

Sk

for

k=2, ..., 6.

S2=

	0	0	81	0	0
	0	0	0	81	0
	0	0	0	0	81
	0	0	0	0	0
	0	0	0	0	0

S3=

	0	0	0	729	0
	0	0	0	0	729
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S4=

	0	0	0	0	6561
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S5=

	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S6=

	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

2.1.47

Let

S=

	0	1	0	0	0
	0	0	1	0	0
	0	0	0	1	0
	0	0	0	0	1
	0	0	0	0	0

.

Compute

Sk

for

k=2, ..., 6.

S2=

	0	0	1	0	0
	0	0	0	1	0
	0	0	0	0	1
	0	0	0	0	0
	0	0	0	0	0

S3=

	0	0	0	1	0
	0	0	0	0	1
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S4=

	0	0	0	0	1
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S5=

	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

S6=

	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0
	0	0	0	0	0

2.2.1

Find the inverse of the matrix.

7	5
8	4

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

	−13	512
	23	−712

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

2.2.1

Find the inverse of the matrix.

3	9
7	2

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

	−257	319
	757	−119

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

2.2.1

Find the inverse of the matrix.

4	8
9	2

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

	−132	18
	964	−116

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

2.2.8

Use the given inverse of the coefficient matrix to solve the following system.

7x1+2x2	=	6	A−1=	1 1 −3 −72
−6x1−2x2	=	−2

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

x1=4

and

x2=−11

(Simplify your answers.)

There is no solution.

2.2.8

Use the given inverse of the coefficient matrix to solve the following system.

7x1+2x2	=	−8	A−1=	1 1 −3 −72
−6x1−2x2	=	4

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

x1=−4

and

x2=10

(Simplify your answers.)

There is no solution.

2.2.9

Let

A=

1	2
8	18

,

b1=

1

16

,

b2=

5

34

,

b3=

2

12

,

and

b4=

6

40

.

a. Find

A−1

and use it solve the four equations

Ax=b1,

Ax=b2

,Ax=b3,

and

Ax=b4.

b. The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A

b1

b2

b3

b4].

a. Find

A−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

The inverse matrix is

A−1=

	9	−1
	−4	12

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

Solve

Ax=b1.

x=

	−7
	4

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b2.

x=

	11
	−3

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b3.

x=

	6
	−2

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b4.

x=

	14
	−4

(Type an integer or simplified fraction for each matrix element.)

b. Solve the four equations by row reducing the augmented matrix [A

b1

b2

b3

b4].

Write the augmented matrix [A

b1

b2

b3

b4]

in reduced echelon form.

	1	0	−7	11	6	14
	0	1	4	−3	−2	−4

(Type an integer or simplified fraction for each matrix element.)

2.2.9

Let

A=

1	2
8	18

,

b1=

−1

−2

,

b2=

5

34

,

b3=

3

22

,

and

b4=

4

22

.

a. Find

A−1

and use it solve the four equations

Ax=b1,

Ax=b2

,Ax=b3,

and

Ax=b4.

b. The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A

b1

b2

b3

b4].

a. Find

A−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

The inverse matrix is

A−1=

	9	−1
	−4	12

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

Solve

Ax=b1.

x=

	−7
	3

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b2.

x=

	11
	−3

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b3.

x=

	5
	−1

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b4.

x=

	14
	−5

(Type an integer or simplified fraction for each matrix element.)

b. Solve the four equations by row reducing the augmented matrix [A

b1

b2

b3

b4].

Write the augmented matrix [A

b1

b2

b3

b4]

in reduced echelon form.

	1	0	−7	11	5	14
	0	1	3	−3	−1	−5

(Type an integer or simplified fraction for each matrix element.)

2.2.9

Let

A=

1	2
7	16

,

b1=

−2

−8

,

b2=

1

−3

,

b3=

−2

−20

,

and

b4=

5

27

.

a. Find

A−1

and use it solve the four equations

Ax=b1,

Ax=b2

,Ax=b3,

and

Ax=b4.

b. The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A

b1

b2

b3

b4].

a. Find

A−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

The inverse matrix is

A−1=

	8	−1
	−72	12

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

Solve

Ax=b1.

x=

	−8
	3

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b2.

x=

	11
	−5

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b3.

x=

	4
	−3

(Type an integer or simplified fraction for each matrix element.)

Solve

Ax=b4.

x=

	13
	−4

(Type an integer or simplified fraction for each matrix element.)

b. Solve the four equations by row reducing the augmented matrix [A

b1

b2

b3

b4].

Write the augmented matrix [A

b1

b2

b3

b4]

in reduced echelon form.

	1	0	−8	11	4	13
	0	1	3	−5	−3	−4

(Type an integer or simplified fraction for each matrix element.)

2.2.10

Use matrix algebra to show that if A is invertible and D satisfies

AD=I,

then

D=A−1.

Choose the correct answer below.

Add

A−1

to both sides of the equation

AD=I

to obtain

A−1+AD=A−1+I,

ID=A−1,

and

D=A−1.

Right-multiply each side of the equation

AD=I

by

A−1

to obtain

ADA−1=IA−1,

DI=A−1,

and

D=A−1.

Left-multiply each side of the equation

AD=I

by

A−1

to obtain

A−1AD=A−1I,

ID=A−1,

and

D=A−1.

Add

A−1

to both sides of the equation

AD=I

to obtain

AD+A−1=I+A−1,

DI=A−1,

and

D=A−1.

2.2.10

Use matrix algebra to show that if A is invertible and D satisfies

AD=I,

then

D=A−1.

Choose the correct answer below.

Right-multiply each side of the equation

AD=I

by

A−1

to obtain

ADA−1=IA−1,

DI=A−1,

and

D=A−1.

Left-multiply each side of the equation

AD=I

by

A−1

to obtain

A−1AD=A−1I,

ID=A−1,

and

D=A−1.

Add

A−1

to both sides of the equation

AD=I

to obtain

A−1+AD=A−1+I,

ID=A−1,

and

D=A−1.

Add

A−1

to both sides of the equation

AD=I

to obtain

AD+A−1=I+A−1,

DI=A−1,

and

D=A−1.

2.2.11

Determine whether the statement below is true or false. Justify the answer.

In order for a matrix B to be the inverse of A, both equations

AB=I

and

BA=I

must be true.

Choose the correct answer below.

The statement is true. The product of a matrix and its inverse is the identity matrix.

The statement is true. Since

AB=BA,

AB=I

if and only if

BA=I.

The statement is false. If

AB=I

and

BC=I,

then A is one inverse of B and C is possibly another inverse of B.

The statement is false. It's possible that the product AB is defined and equals

I,

yet the product BA is not defined.

2.2.11

Determine whether the statement below is true or false. Justify the answer.

In order for a matrix B to be the inverse of A, both equations

AB=I

and

BA=I

must be true.

Choose the correct answer below.

The statement is false. If

AB=I

and

BC=I,

then A is one inverse of B and C is possibly another inverse of B.

The statement is false. It's possible that the product AB is defined and equals

I,

yet the product BA is not defined.

The statement is true. The product of a matrix and its inverse is the identity matrix.

The statement is true. Since

AB=BA,

AB=I

if and only if

BA=I.

2.2.12

Determine whether the statement below is true or false. Justify the answer.

A product of invertible

n×n

matrices is invertible, and the inverse of the product is the product of their inverses in the same order.

Choose the correct answer below.

The statement is true. If A and B are invertible matrices, then

(AB)−1=A−1B−1.

The statement is false. If A and B are invertible matrices, then

(AB)−1=B−1A−1.

The statement is true. Since invertible matrices commute,

(AB)−1=B−1A−1=A−1B−1.

The statement is false. If A and B are invertible matrices, then

(AB)−1=BA−1B−1.

2.2.13

Determine whether the statement below is true or false. Justify the answer.

If A and B are

n×n

and invertible, then

A−1B−1

is the inverse of AB.

Choose the correct answer below.

The statement is false. The inverse of AB is

1AB.

The statement is false. The inverse of AB is

B−1A−1.

The statement is true. The following equalities hold,

(AB)−1=(BA)−1=A−1B−1.

The statement is true. The following equalities hold,

A−1B−1=B−1A−1=(AB)−1.

2.2.13

Determine whether the statement below is true or false. Justify the answer.

If A and B are

n×n

and invertible, then

A−1B−1

is the inverse of AB.

Choose the correct answer below.

The statement is false. The inverse of AB is

B−1A−1.

The statement is true. The following equalities hold,

(AB)−1=(BA)−1=A−1B−1.

The statement is true. The following equalities hold,

A−1B−1=B−1A−1=(AB)−1.

The statement is false. The inverse of AB is

1AB.

2.2.14

Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then the inverse of

A−1

is A itself.

Choose the correct answer below.

The statement is true. A is invertible if and only if

A=A−1.

Since A is invertible,

A=A−1.

Since

A=A−1,

A−1

is invertible, and A is the inverse of

A−1.

The statement is false. It does not follow from the fact that A is invertible that

A−1

is also invertible.

The statement is false. Since inverses are not unique, it is possible that

B≠A

is the inverse of

A−1.

The statement is true. Since

A−1

is the inverse of A,

A−1A=I=AA−1.

Since

A−1A=I=AA−1,

A is the inverse of

A−1.

2.2.14

Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then the inverse of

A−1

is A itself.

Choose the correct answer below.

The statement is true. Since

A−1

is the inverse of A,

A−1A=I=AA−1.

Since

A−1A=I=AA−1,

A is the inverse of

A−1.

The statement is false. It does not follow from the fact that A is invertible that

A−1

is also invertible.

The statement is true. A is invertible if and only if

A=A−1.

Since A is invertible,

A=A−1.

Since

A=A−1,

A−1

is invertible, and A is the inverse of

A−1.

The statement is false. Since inverses are not unique, it is possible that

B≠A

is the inverse of

A−1.

2.2.14

Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then the inverse of

A−1

is A itself.

Choose the correct answer below.

The statement is true. Since

A−1

is the inverse of A,

A−1A=I=AA−1.

Since

A−1A=I=AA−1,

A is the inverse of

A−1.

The statement is false. It does not follow from the fact that A is invertible that

A−1

is also invertible.

The statement is false. Since inverses are not unique, it is possible that

B≠A

is the inverse of

A−1.

The statement is true. A is invertible if and only if

A=A−1.

Since A is invertible,

A=A−1.

Since

A=A−1,

A−1

is invertible, and A is the inverse of

A−1.

2.2.14

Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then the inverse of

A−1

is A itself.

Choose the correct answer below.

The statement is false. Since inverses are not unique, it is possible that

B≠A

is the inverse of

A−1.

The statement is true. A is invertible if and only if

A=A−1.

Since A is invertible,

A=A−1.

Since

A=A−1,

A−1

is invertible, and A is the inverse of

A−1.

The statement is false. It does not follow from the fact that A is invertible that

A−1

is also invertible.

The statement is true. Since

A−1

is the inverse of A,

A−1A=I=AA−1.

Since

A−1A=I=AA−1,

A is the inverse of

A−1.

2.2.14

Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then the inverse of

A−1

is A itself.

Choose the correct answer below.

The statement is false. Since inverses are not unique, it is possible that

B≠A

is the inverse of

A−1.

The statement is true. A is invertible if and only if

A=A−1.

Since A is invertible,

A=A−1.

Since

A=A−1,

A−1

is invertible, and A is the inverse of

A−1.

The statement is false. It does not follow from the fact that A is invertible that

A−1

is also invertible.

The statement is true. Since

A−1

is the inverse of A,

A−1A=I=AA−1.

Since

A−1A=I=AA−1,

A is the inverse of

A−1.

2.2.15

Determine whether the statement below is true or false. Justify the answer.

If

A=

	a	b
	c	d

and

ab−cd≠0,

then A is invertible.

Choose the correct answer below.

The statement is false. If A is invertible, then

ab=cd.

The statement is true. The matrix

A=

	a	b
	c	d

is invertible if and only if

a≠c

and

b≠d.

The statement is true. The inverse of A is

A−1=1ab−cd

	d	−b
	−c	a

.

This expression is always defined when

ab−cd≠0.

The statement is false. If

ad−bc≠0,

then A is invertible.

2.2.15

Determine whether the statement below is true or false. Justify the answer.

If

A=

	a	b
	c	d

and

ab−cd≠0,

then A is invertible.

Choose the correct answer below.

The statement is true. The inverse of A is

A−1=1ab−cd

	d	−b
	−c	a

.

This expression is always defined when

ab−cd≠0.

The statement is false. If

ad−bc≠0,

then A is invertible.

The statement is false. If A is invertible, then

ab=cd.

The statement is true. The matrix

A=

	a	b
	c	d

is invertible if and only if

a≠c

and

b≠d.

2.2.16

Determine whether the statement below is true or false. Justify the answer.

If

A=

	a	b
	c	d

and

ad=bc,

then A is not invertible.

Choose the correct answer below.

The statement is false. If A is invertible, then

ad=bc.

The statement is false. If

ad=bc,

then A is invertible.

The statement is true. If

ad=bc

then

ad−bc=0,

and

1ad−bc

	d	−b
	−c	a

is undefined.

The statement is true. The matrix

A=

	a	b
	c	d

is invertible if and only if

a≠b

and

b≠d.

2.2.16

Determine whether the statement below is true or false. Justify the answer.

If

A=

	a	b
	c	d

and

ad=bc,

then A is not invertible.

Choose the correct answer below.

The statement is true. The matrix

A=

	a	b
	c	d

is invertible if and only if

a≠b

and

b≠d.

The statement is true. If

ad=bc

then

ad−bc=0,

and

1ad−bc

	d	−b
	−c	a

is undefined.

The statement is false. If

ad=bc,

then A is invertible.

The statement is false. If A is invertible, then

ad=bc.

2.2.17

Determine whether the statement below is true or false. Justify the answer.

If A is an invertible

n×n

matrix, then the equation

Ax=b

is consistent for each

b

in

ℝn.

Choose the correct answer below.

The statement is false. The matrix A satisfies

Ax=b

if and only if A is row equivalent to the identity matrix, and not every matrix that is row equivalent to the identity matrix is invertible.

The statement is true. Since A is invertible,

A−1b

exists for all

b

in

ℝn.

Define

x=A−1b.

Then

Ax=b.

The statement is true. Since A is invertible,

A−1b=x

for all

x

in

ℝn.

Multiply both sides by A and the result is

Ax=b.

The statement is false. The matrix A is invertible if and only if A is row equivalent to the identity matrix, and not every matrix A satisfying

Ax=b

is row equivalent to the identity matrix.

2.2.17

Determine whether the statement below is true or false. Justify the answer.

If A is an invertible

n×n

matrix, then the equation

Ax=b

is consistent for each

b

in

ℝn.

Choose the correct answer below.

The statement is false. The matrix A is invertible if and only if A is row equivalent to the identity matrix, and not every matrix A satisfying

Ax=b

is row equivalent to the identity matrix.

The statement is true. Since A is invertible,

A−1b=x

for all

x

in

ℝn.

Multiply both sides by A and the result is

Ax=b.

The statement is false. The matrix A satisfies

Ax=b

if and only if A is row equivalent to the identity matrix, and not every matrix that is row equivalent to the identity matrix is invertible.

The statement is true. Since A is invertible,

A−1b

exists for all

b

in

ℝn.

Define

x=A−1b.

Then

Ax=b.

2.2.17

Determine whether the statement below is true or false. Justify the answer.

If A is an invertible

n×n

matrix, then the equation

Ax=b

is consistent for each

b

in

ℝn.

Choose the correct answer below.

The statement is true. Since A is invertible,

A−1b=x

for all

x

in

ℝn.

Multiply both sides by A and the result is

Ax=b.

The statement is false. The matrix A satisfies

Ax=b

if and only if A is row equivalent to the identity matrix, and not every matrix that is row equivalent to the identity matrix is invertible.

The statement is true. Since A is invertible,

A−1b

exists for all

b

in

ℝn.

Define

x=A−1b.

Then

Ax=b.

The statement is false. The matrix A is invertible if and only if A is row equivalent to the identity matrix, and not every matrix A satisfying

Ax=b

is row equivalent to the identity matrix.

2.2.17

Determine whether the statement below is true or false. Justify the answer.

If A is an invertible

n×n

matrix, then the equation

Ax=b

is consistent for each

b

in

ℝn.

Choose the correct answer below.

The statement is true. Since A is invertible,

A−1b=x

for all

x

in

ℝn.

Multiply both sides by A and the result is

Ax=b.

The statement is false. The matrix A is invertible if and only if A is row equivalent to the identity matrix, and not every matrix A satisfying

Ax=b

is row equivalent to the identity matrix.

The statement is true. Since A is invertible,

A−1b

exists for all

b

in

ℝn.

Define

x=A−1b.

Then

Ax=b.

The statement is false. The matrix A satisfies

Ax=b

if and only if A is row equivalent to the identity matrix, and not every matrix that is row equivalent to the identity matrix is invertible.

2.2.18

Determine whether the statement below is true or false. Justify the answer.

If A can be row reduced to the identity matrix, then A must be invertible.

Choose the correct answer below.

The statement is false. Since the identity matrix is not invertible, A is not invertible either.

The statement is true. Since A can be row reduced to the identity matrix, A is row equivalent to the identity matrix. Since every matrix that is row equivalent to the identity is invertible, A is invertible.

The statement is false. Not every matrix that is row equivalent to the identity matrix is invertible.

The statement is true. Since A can be row reduced to the identity matrix, I is the inverse of A. Therefore, A is invertible.

2.2.18

Determine whether the statement below is true or false. Justify the answer.

If A can be row reduced to the identity matrix, then A must be invertible.

Choose the correct answer below.

The statement is true. Since A can be row reduced to the identity matrix, A is row equivalent to the identity matrix. Since every matrix that is row equivalent to the identity is invertible, A is invertible.

The statement is false. Not every matrix that is row equivalent to the identity matrix is invertible.

The statement is false. Since the identity matrix is not invertible, A is not invertible either.

The statement is true. Since A can be row reduced to the identity matrix, I is the inverse of A. Therefore, A is invertible.

2.2.39

Find the inverse of the matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	6
4	23

Set up the matrix

[A I].

[A I]=

	1	6	1	0
	4	23	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	−23	6
	4	−1

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.39

Find the inverse of the matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	3
4	11

Set up the matrix

[A I].

[A I]=

	1	3	1	0
	4	11	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	−11	3
	4	−1

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.40

Find the inverse of the matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A  I].

A=

4	3
3	5

Set up the matrix

[A  I].

[A  I]=

	4	3	1	0
	3	5	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	511	−311
	−311	411

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

2.2.40

Find the inverse of the matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A  I].

A=

6	3
4	7

Set up the matrix

[A  I].

[A  I]=

	6	3	1	0
	4	7	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	730	−110
	−215	15

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

2.2.41

Find the inverse of the given matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	0	2
−3	1	3
−4	4	3

Set up the matrix

[A I].

[A I]=

	1	0	2	1	0	0
	−3	1	3	0	1	0
	−4	4	3	0	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	925	−825	225
	325	−1125	925
	825	425	−125

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.41

Find the inverse of the given matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	0	−3
−2	1	−4
4	2	−4

Set up the matrix

[A I].

[A I]=

	1	0	−3	1	0	0
	−2	1	−4	0	1	0
	4	2	−4	0	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	17	−314	328
	−67	27	514
	−27	−114	128

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.41

Find the inverse of the given matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	0	−3
−2	1	3
2	−4	3

Set up the matrix

[A I].

[A I]=

	1	0	−3	1	0	0
	−2	1	3	0	1	0
	2	−4	3	0	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	−5	−4	−1
	−4	−3	−1
	−2	−43	−13

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.41

Find the inverse of the given matrix, if it exists. Use the algorithm for finding

A−1

by row reducing

[A I].

A=

1	0	3
−2	1	2
−4	4	2

Set up the matrix

[A I].

[A I]=

	1	0	3	1	0	0
	−2	1	2	0	1	0
	−4	4	2	0	0	1

(Type an integer or simplified fraction for each matrix element.)

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	13	−23	16
	29	−79	49
	29	29	−118

.

(Type an integer or simplified fraction for each matrix element.)

The matrix A does not have an inverse.

2.2.42

Find the inverse of the given matrix, if it exists.

A=

1	−2	1
2	−3	1
−4	6	−2

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A−1=

(Type an integer or decimal for each matrix element.)

The matrix A does not have an inverse.

2.2.42

Find the inverse of the given matrix, if it exists.

A=

1	−2	1
−4	9	−5
2	−6	4

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A−1=

(Type an integer or decimal for each matrix element.)

The matrix A does not have an inverse.

2.2.42

Find the inverse of the given matrix, if it exists.

A=

1	−2	1
2	−3	1
4	−2	−2

Find the inverse. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A−1=

(Type an integer or decimal for each matrix element.)

The matrix A does not have an inverse.

2.2.47

Let

A=

	1	5
	1	6
	1	11

.

Construct a

2×3

matrix C (by trial and error) using only 1,

−1,

and 0 as entries, such that

CA=I2.

Compute AC and note that

AC≠I3.

C=

	1	1	−1
	−1	1	0

AC=

	−4	6	−1
	−5	7	−1
	−10	12	−1

(Simplify your answer.)

2.3.3

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

2	0	0
−3	−6	0
7	5	−3

Choose the correct answer below.

The matrix is not invertible. If the given matrix is A, the equation

Ax=b

has no solution for some b in

ℝ3.

The matrix is not invertible. The given matrix has two pivot positions.

The matrix is invertible. The given matrix has three pivot positions.

The matrix is invertible. If the given matrix is A, there is a

3×3

matrix C such that

CI=A.

2.3.4

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

3	0	−3
2	0	3
−3	0	8

Choose the correct answer below.

The matrix is invertible. The given matrix has 2 pivot positions.

The matrix is not invertible. If the given matrix is A, the equation

Ax=0

has only the trivial solution.

The matrix is invertible. The columns of the given matrix span

ℝ3.

The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.

2.3.4

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

4	0	−4
3	0	5
−3	0	7

Choose the correct answer below.

The matrix is invertible. The columns of the given matrix span

ℝ3.

The matrix is not invertible. If the given matrix is A, the equation

Ax=0

has only the trivial solution.

The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.

The matrix is invertible. The given matrix has 2 pivot positions.

2.3.4

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

4	0	−5
3	0	7
−4	0	8

Choose the correct answer below.

The matrix is invertible. The given matrix has 2 pivot positions.

The matrix is invertible. The columns of the given matrix span

ℝ3.

The matrix is not invertible. If the given matrix is A, the columns of A do not form a linearly independent set.

The matrix is not invertible. If the given matrix is A, the equation

Ax=0

has only the trivial solution.

2.3.5

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

0	4	−5
1	0	3
−3	−8	1

Choose the correct answer below.

The matrix is not invertible. If the given matrix is A, the equation

Ax=b

has a solution for at least one b in

ℝ3.

The matrix is not invertible. If the given matrix is A, A is not row equivalent to the

n×n

identity matrix.

The matrix is invertible. The columns of the given matrix span are linearly dependent.

The matrix is invertible. The given matrix has 3 pivot positions.

2.3.8

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

4	5	7	5
0	1	4	6
0	0	2	8
0	0	0	1

Choose the correct answer below.

The matrix is not invertible. If the given matrix is A, the equation

Ax=b

has no solution for at least one b in

ℝ4.

The matrix is invertible. The given matrix has 4 pivot positions.

The matrix is not invertible. In the given matrix the columns do not form a linearly independent set.

The matrix is invertible. The given matrix is not row equivalent to the

n×n

identity matrix.

2.3.8

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer.

3	5	8	5
0	1	3	6
0	0	2	9
0	0	0	1

Choose the correct answer below.

The matrix is not invertible. If the given matrix is A, the equation

Ax=b

has no solution for at least one b in

ℝ4.

The matrix is invertible. The given matrix has 4 pivot positions.

The matrix is not invertible. In the given matrix the columns do not form a linearly independent set.

The matrix is invertible. The given matrix is not row equivalent to the

n×n

identity matrix.

2.3.11

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If the equation

Ax=0

has only the trivial solution, then A is row equivalent to the

n×n

identity matrix.

Choose the correct answer below.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible. Thus, A cannot be row equivalent to the

n×n

identity matrix.

The statement is true. By the Invertible Matrix Theorem, if equation

Ax=0

has only the trivial solution, then the equation

Ax=b

has no solutions for each b in

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible; this means the columns of A do not span

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is true. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the

n×n

identity matrix.

2.3.11

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If the equation

Ax=0

has only the trivial solution, then A is row equivalent to the

n×n

identity matrix.

Choose the correct answer below.

The statement is true. By the Invertible Matrix Theorem, if equation

Ax=0

has only the trivial solution, then the equation

Ax=b

has no solutions for each b in

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible. Thus, A cannot be row equivalent to the

n×n

identity matrix.

The statement is true. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible; this means the columns of A do not span

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

2.3.11

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If the equation

Ax=0

has only the trivial solution, then A is row equivalent to the

n×n

identity matrix.

Choose the correct answer below.

The statement is true. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible. Thus, A cannot be row equivalent to the

n×n

identity matrix.

The statement is true. By the Invertible Matrix Theorem, if equation

Ax=0

has only the trivial solution, then the equation

Ax=b

has no solutions for each b in

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

The statement is false. By the Invertible Matrix Theorem, if the equation

Ax=0

has only the trivial solution, then the matrix is not invertible; this means the columns of A do not span

ℝn.

Thus, A must also be row equivalent to the

n×n

identity matrix.

2.3.12

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If there is an

n×n

matrix D such that

AD=I,

then there is also an

n×n

matrix C such that

CA=I.

Choose the correct answer below.

The statement is false. It does not follow from the Invertible Matrix Theorem that if

AD=I,

then

CA=I.

The statement is true. By the Invertible Matrix Theorem, if there is an

n×n

matrix D such that

AD=I,

then it must be true that there is also an

n×n

matrix C such that

CA=I.

The statement is false. Matrix multiplication is not commutative. It is possible that only D (and not A) is invertible in the equation

AD=I.

This implies that

CA=I

is only true when C is invertible (for cases where A is not invertible), but it is not given that C is invertible.

The statement is true. By the Invertible Matrix Theorem, if

AD=I

then A, D, or both is/are the identity matrix. Therefore,

CA=I.

2.3.12

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If there is an

n×n

matrix D such that

AD=I,

then there is also an

n×n

matrix C such that

CA=I.

Choose the correct answer below.

The statement is false. Matrix multiplication is not commutative. It is possible that only D (and not A) is invertible in the equation

AD=I.

This implies that

CA=I

is only true when C is invertible (for cases where A is not invertible), but it is not given that C is invertible.

The statement is true. By the Invertible Matrix Theorem, if

AD=I

then A, D, or both is/are the identity matrix. Therefore,

CA=I.

The statement is false. It does not follow from the Invertible Matrix Theorem that if

AD=I,

then

CA=I.

The statement is true. By the Invertible Matrix Theorem, if there is an

n×n

matrix D such that

AD=I,

then it must be true that there is also an

n×n

matrix C such that

CA=I.

2.3.15

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If A is an

n×n

matrix, then the equation

Ax=b

has at least one solution for each

b

in

ℝn.

Choose the correct answer below.

The statement is true. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

for all matrices of size

n×n.

The statement is false. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

only if a matrix is invertible.

The statement is true. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the matrix is not invertible.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the equation

Ax=b

has no solution.

2.3.15

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If A is an

n×n

matrix, then the equation

Ax=b

has at least one solution for each

b

in

ℝn.

Choose the correct answer below.

The statement is true. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the matrix is not invertible.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the equation

Ax=b

has no solution.

The statement is true. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

for all matrices of size

n×n.

The statement is false. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

only if a matrix is invertible.

2.3.15

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If A is an

n×n

matrix, then the equation

Ax=b

has at least one solution for each

b

in

ℝn.

Choose the correct answer below.

The statement is true. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the matrix is not invertible.

The statement is true. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

for all matrices of size

n×n.

The statement is false. By the Invertible Matrix Theorem,

Ax=b

has at least one solution for each

b

in

ℝn

only if a matrix is invertible.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each

b

in

ℝn,

then the equation

Ax=b

has no solution.

2.3.16

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If the equation

Ax=b

has at least one solution for each b in

ℝn,

then the solution is unique for each

b.

Choose the correct answer below.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then matrix A is not invertible. If A is not invertible, then according to the Invertible Matrix Theorem, the solution is not unique for each

b.

The statement is true. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then matrix A is invertible. If A is invertible, then according to the invertible matrix theorem the solution is unique for each

b.

The statement is true, but only for

x≠0.

By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then the equation

Ax=0

does not only have the trivial solution.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then the linear transformation

x ↦ Ax

does not map

ℝn

onto

ℝn.

2.3.16

For this exercise assume that the matrices are all

n×n.

The statement in this exercise is an implication of the form "If "statement 1", then "statement 2"." Mark an implication as True if the truth of "statement 2" always follows whenever "statement 1" happens to be true. Mark the implication as False if "statement 2" is false but "statement 1" is true. Justify your answer.

If the equation

Ax=b

has at least one solution for each b in

ℝn,

then the solution is unique for each

b.

Choose the correct answer below.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then the linear transformation

x ↦ Ax

does not map

ℝn

onto

ℝn.

The statement is true. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then matrix A is invertible. If A is invertible, then according to the invertible matrix theorem the solution is unique for each

b.

The statement is true, but only for

x≠0.

By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then the equation

Ax=0

does not only have the trivial solution.

The statement is false. By the Invertible Matrix Theorem, if

Ax=b

has at least one solution for each b in

ℝn,

then matrix A is not invertible. If A is not invertible, then according to the Invertible Matrix Theorem, the solution is not unique for each

b.

2.3.21

An m×n upper triangular matrix is one whose entries below the main diagonal are zeros, as is shown in the matrix to the right. When is a square upper triangular matrix invertible? Justify your answer.

3 4 7 4 0 1 4 6 0 0 2 8 0 0 0 1

Choose the correct answer below.

A square upper triangular matrix is invertible when all entries on the main diagonal are ones. If any entry on the main diagonal is not one, then the equation

Ax=b,

where A is an

n×n

square upper triangular matrix, has no solution for at least one b in

ℝn.

A square upper triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the

n×n

matrix has n pivot positions.

A square upper triangular matrix is invertible when all entries above the main diagonal are zeros as well. This means that the matrix is row equivalent to the

n×n

identity matrix.

A square upper triangular matrix is invertible when the matrix is equal to its own transpose. For such a matrix A,

A=AT

means that the equation

Ax=b

has at least one solution for each b in

ℝn.

2.3.21

An m×n upper triangular matrix is one whose entries below the main diagonal are zeros, as is shown in the matrix to the right. When is a square upper triangular matrix invertible? Justify your answer.

3 4 7 4 0 1 4 6 0 0 2 8 0 0 0 1

Choose the correct answer below.

A square upper triangular matrix is invertible when the matrix is equal to its own transpose. For such a matrix A,

A=AT

means that the equation

Ax=b

has at least one solution for each b in

ℝn.

A square upper triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the

n×n

matrix has n pivot positions.

A square upper triangular matrix is invertible when all entries on the main diagonal are ones. If any entry on the main diagonal is not one, then the equation

Ax=b,

where A is an

n×n

square upper triangular matrix, has no solution for at least one b in

ℝn.

A square upper triangular matrix is invertible when all entries above the main diagonal are zeros as well. This means that the matrix is row equivalent to the

n×n

identity matrix.

2.3.21

An m×n upper triangular matrix is one whose entries below the main diagonal are zeros, as is shown in the matrix to the right. When is a square upper triangular matrix invertible? Justify your answer.

3 4 7 4 0 1 4 6 0 0 2 8 0 0 0 1

Choose the correct answer below.

A square upper triangular matrix is invertible when all entries on the main diagonal are ones. If any entry on the main diagonal is not one, then the equation

Ax=b,

where A is an

n×n

square upper triangular matrix, has no solution for at least one b in

ℝn.

A square upper triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the

n×n

matrix has n pivot positions.

A square upper triangular matrix is invertible when all entries above the main diagonal are zeros as well. This means that the matrix is row equivalent to the

n×n

identity matrix.

A square upper triangular matrix is invertible when the matrix is equal to its own transpose. For such a matrix A,

A=AT

means that the equation

Ax=b

has at least one solution for each b in

ℝn.

2.3.22

An m×n lower triangular matrix is one whose entries above the main diagonal are zeros, as is shown in the matrix to the right. When is a square lower triangular matrix invertible? Justify your answer.

3 0 0 0 4 1 0 0 7 4 2 0 4 6 8 1

Choose the correct answer below.

A square lower triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the

n×n

matrix has n pivot positions.

A square lower triangular matrix is invertible when all entries below the main diagonal are zeros as well. This means that the matrix is row equivalent to the

n×n

identity matrix.

A square lower triangular matrix is invertible when all entries on the main diagonal are ones. If any entry on the main diagonal is not one, then the equation

Ax=b,

where A is an

n×n

square lower triangular matrix, has no solution for at least one b in

ℝn.

A square lower triangular matrix is invertible when the matrix is equal to its own transpose. For such a matrix A,

A=AT

means that the equation

Ax=b

has at least one solution for each b in

ℝn.

2.3.22

An m×n lower triangular matrix is one whose entries above the main diagonal are zeros, as is shown in the matrix to the right. When is a square lower triangular matrix invertible? Justify your answer.

3 0 0 0 4 1 0 0 7 4 2 0 4 6 8 1

Choose the correct answer below.

A square lower triangular matrix is invertible when the matrix is equal to its own transpose. For such a matrix A,

A=AT

means that the equation

Ax=b

has at least one solution for each b in

ℝn.

A square lower triangular matrix is invertible when all entries on its main diagonal are nonzero. If all of the entries on its main diagonal are nonzero, then the

n×n

matrix has n pivot positions.

A square lower triangular matrix is invertible when all entries below the main diagonal are zeros as well. This means that the matrix is row equivalent to the

n×n

identity matrix.

A square lower triangular matrix is invertible when all entries on the main diagonal are ones. If any entry on the main diagonal is not one, then the equation

Ax=b,

where A is an

n×n

square lower triangular matrix, has no solution for at least one b in

ℝn.

2.3.42

The given T is a linear transformation from

ℝ2

into

ℝ2.

Show that T is invertible and find a formula for

T−1.

Tx1,x2=4x1−6x2,−4x1+9x2

To show that T is invertible, calculate the determinant of the standard matrix for T. The determinant of the standard matrix is

12.

(Simplify your answer.)

T−1x1,x2=34x1+12x2,13x1+13x2

(Type an ordered pair. Type an expression using

x1

and

x2

as the variables.)

2.8.1

A set in ℝ2 is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of ℝ2. (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice.

The set is not a subspace because it does not include the zero vector.

0

The set is not a subspace because it is not closed under either scalar multiplication or sums. For example,

multiplied by (1,3) is not in the set, and the sum of (3,1) and (1,3) is not in the set.

u

v

u+v

ku

The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (3,1) and (1,3) is not in the set.

u

v

u+v

The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example,

−1

multiplied by (1,1) is not in the set.

u

ku

2.8.1

A set in ℝ2 is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of ℝ2. (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice.

The set is not a subspace because it does not include the zero vector.

0

The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example,

−1

multiplied by (1,1) is not in the set.

u

ku

The set is not a subspace because it is not closed under either scalar multiplication or sums. For example,

multiplied by (1,3) is not in the set, and the sum of (3,1) and (1,3) is not in the set.

u

v

u+v

ku

The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (3,1) and (1,3) is not in the set.

u

v

u+v

2.8.2

A set in ℝ2 is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of ℝ2. (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice.

The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (2,2) and

(−1,−3)

is not in the set.

u

v

u+v

The set is not a subspace because it does not include the zero vector.

0

The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example,

multiplied by (0,1) is not in the set.

u

ku

The set is not a subspace because it is not closed under either scalar multiplication or sums. For example,

multiplied by (0,1) is not in the set, and the sum of (2,2) and

(−1,−3)

is not in the set.

u

v

u+v

ku

2.8.2

A set in ℝ2 is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of ℝ2. (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice.

The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example,

multiplied by (0,1) is not in the set.

u

ku

The set is not a subspace because it is not closed under either scalar multiplication or sums. For example,

multiplied by (0,1) is not in the set, and the sum of (2,2) and

(−1,−3)

is not in the set.

u

v

u+v

ku

The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (2,2) and

(−1,−3)

is not in the set.

u

v

u+v

The set is not a subspace because it does not include the zero vector.

0

2.8.5

Let

v1=

1

3

−5

,

v2=

−2

−3

9

,

and

w=

−3

−3

13

.

Determine if

w

is

in the subspace of

ℝ3

generated by

v1

and

v2.

Is

w

is

in the subspace of

ℝ3

generated by

v1

and

v2?

Yes

No

2.8.5

Let

v1=

1

3

−4

,

v2=

−2

−3

7

,

and

w=

−3

−3

10

.

Determine if

w

is

in the subspace of

ℝ3

generated by

v1

and

v2.

Is

w

is

in the subspace of

ℝ3

generated by

v1

and

v2?

No

Yes

2.8.5

Let

v1=

1

3

−5

,

v2=

−2

−2

9

,

and

w=

−3

−1

13

.

Determine if

w

is

in the subspace of

ℝ3

generated by

v1

and

v2.

Is

w

is

in the subspace of

ℝ3

generated by

v1

and

v2?

No

Yes

2.8.7

Let

v1=

	1
	−3
	0

,

v2=

	0
	1
	4

,

v3=

	4
	−6
	24

,

p=

	2
	−3
	12

,

and

A=

	v1	v2	v3

.

a. How many vectors are in

{v1,

v2,

v3}?

b. How many vectors are in Col A?

c. Is

p

in Col A? Why or why not?

a. How many vectors are in

{v1,

v2,

v3}?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

3

(Type a whole number.)

There are infinitely many vectors in

{v1,

v2,

v3}.

b. How many vectors are in Col A? Select the correct choice below and, if necessary, fill in the answer box within your choice.

(Type a whole number.)

There are infinitely many vectors in Col A.

c. Is

p

in Col A? Why or why not?

p

is in Col A because A has pivot positions in every row.

p

is in Col A because the system

	A	p

is consistent.

p

is not in Col A because A has too few pivot positions.

p

is not in Col A because the system

	A	p

is not consistent.

2.8.7

Let

v1=

	1
	−2
	0

,

v2=

	0
	1
	2

,

v3=

	4
	−5
	6

,

p=

	4
	−2
	12

,

and

A=

	v1	v2	v3

.

a. How many vectors are in

{v1,

v2,

v3}?

b. How many vectors are in Col A?

c. Is

p

in Col A? Why or why not?

a. How many vectors are in

{v1,

v2,

v3}?

Select the correct choice below and, if necessary, fill in the answer box within your choice.

3

(Type a whole number.)

There are infinitely many vectors in

{v1,

v2,

v3}.

b. How many vectors are in Col A? Select the correct choice below and, if necessary, fill in the answer box within your choice.

(Type a whole number.)

There are infinitely many vectors in Col A.

c. Is

p

in Col A? Why or why not?

p

is in Col A because the system

	A	p

is consistent.

p

is not in Col A because A has too few pivot positions.

p

is not in Col A because the system

	A	p

is not consistent.

p

is in Col A because A has pivot positions in every row.

2.8.9

Let

v1=

	−9
	0
	18

,

v2=

	−6
	6
	9

,

v3=

	0
	−18
	9

,

and

p=

	2
	10
	−7

.

Determine if

p

is in Nul A, where

A=

	v1	v2	v3

.

Is

p

in Nul A?

No, because

Ap

is not equal to the zero vector.

Yes, because

Ap

is equal to the zero vector.

No, because the augmented matrix

	A	p

is not consistent.

Yes, because the augmented matrix

	A	p

is consistent.

2.8.9

Let

v1=

	−3
	0
	6

,

v2=

	−2
	2
	3

,

v3=

	0
	−6
	3

,

and

p=

	5
	6
	−8

.

Determine if

p

is in Nul A, where

A=

	v1	v2	v3

.

Is

p

in Nul A?

No, because

Ap

is not equal to the zero vector.

No, because the augmented matrix

	A	p

is not consistent.

Yes, because the augmented matrix

	A	p

is consistent.

Yes, because

Ap

is equal to the zero vector.

2.8.9

Let

v1=

	−6
	0
	12

,

v2=

	−4
	4
	6

,

v3=

	0
	−12
	6

,

and

p=

	5
	8
	−1

.

Determine if

p

is in Nul A, where

A=

	v1	v2	v3

.

Is

p

in Nul A?

No, because

Ap

is not equal to the zero vector.

No, because the augmented matrix

	A	p

is not consistent.

Yes, because

Ap

is equal to the zero vector.

Yes, because the augmented matrix

	A	p

is consistent.

2.8.10

Let

v1=

−4

1

5

,

v2=

−4

5

5

,

v3=

1

−7

7

,

and

u=

−7

7

5

.

Determine if u is in Nul A, where

A=

v1

v2

v3

.

Is u in Nul A?

No

Yes

2.8.10

Let

v1=

−3

0

4

,

v2=

−3

2

2

,

v3=

0

−7

7

,

and

u=

−7

7

2

.

Determine if u is in Nul A, where

A=

v1

v2

v3

.

Is u in Nul A?

No

Yes

2.8.10

Let

v1=

−3

0

12

,

v2=

−3

6

6

,

v3=

0

−8

8

,

and

u=

−8

8

6

.

Determine if u is in Nul A, where

A=

v1

v2

v3

.

Is u in Nul A?

Yes

No

2.8.11

Give integers p and q such that

Nul A

is a subspace of

ℝp

and

Col A

is a subspace of

ℝq.

A=

1	2	3
4	5	7

Nul A

is a subspace of

ℝp

for

p=3

and

Col A

is a subspace of

ℝq

for

q=2.

2.8.11

Give integers p and q such that

Nul A

is a subspace of

ℝp

and

Col A

is a subspace of

ℝq.

A=

−5	4	3	−2	1
0	3	−4	0	8
9	0	6	−6	1

Nul A

is a subspace of

ℝp

for

p=5

and

Col A

is a subspace of

ℝq

for

q=3.

2.8.13

For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.

A=

	3	2	−1	−2
	−12	−6	12	0
	9	0	−27	18

Find a nonzero vector in Nul A.

	1
	0
	1
	1

Find a nonzero vector in Col A.

	3
	−12
	9

2.8.15

Determine if the set is a basis for

ℝ2.

Justify your answer.

−16

4

,

−64

−2

Is the given set a basis for

ℝ2?

No, because these vectors form the columns of an invertible

2×2

matrix. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these vectors form the columns of an invertible

2×2

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these vectors form the columns of a

2×2

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these vectors form the columns of a

2×2

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.15

Determine if the set is a basis for

ℝ2.

Justify your answer.

−8

2

,

−32

9

Is the given set a basis for

ℝ2?

No, because these vectors form the columns of a

2×2

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these vectors form the columns of an invertible

2×2

matrix. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these vectors form the columns of an invertible

2×2

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these vectors form the columns of a

2×2

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.17

Determine if the set is a basis for

ℝ3.

Justify your answer.

0

0

−4

,

3

6

8

,

4

8

4

Is the given set a basis for

ℝ3?

No, because these three vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes,

because these three vectors form the columns of an invertible 3×3 matrix.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these three vectors form the columns of a

3×3

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No,

because these three vectors form the columns of a 3×3 matrix that is not invertible.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.17

Determine if the set is a basis for

ℝ3.

Justify your answer.

0

0

−2

,

1

2

4

,

2

4

2

Is the given set a basis for

ℝ3?

Yes, because these three vectors form the columns of a

3×3

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes,

because these three vectors form the columns of an invertible 3×3 matrix.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No,

because these three vectors form the columns of a 3×3 matrix that is not invertible.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these three vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.17

Determine if the set is a basis for

ℝ3.

Justify your answer.

0

0

−2

,

−7

0

4

,

3

3

2

Is the given set a basis for

ℝ3?

Yes, because these three vectors form the columns of a

3×3

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes,

because these three vectors form the columns of an invertible 3×3 matrix.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No,

because these three vectors form the columns of a 3×3 matrix that is not invertible.

By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these three vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.19

Determine if the set is a basis for

ℝ3.

Justify your answer.

3

−4

1

,

6

1

−4

Is the given set a basis for

ℝ3?

Yes, because these vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these two vectors are linearly dependent.

No, because these vectors form a matrix with only 2 pivot columns. Therefore, these vectors form a basis for a two-dimensional subspace of

ℝ3.

Yes, because these two vectors are linearly independent.

2.8.19

Determine if the set is a basis for

ℝ3.

Justify your answer.

4

−4

1

,

8

1

−4

Is the given set a basis for

ℝ3?

No, because these vectors form a matrix with only 2 pivot columns. Therefore, these vectors form a basis for a two-dimensional subspace of

ℝ3.

Yes, because these two vectors are linearly independent.

Yes, because these vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these two vectors are linearly dependent.

2.8.19

Determine if the set is a basis for

ℝ3.

Justify your answer.

2

−8

1

,

4

2

−5

Is the given set a basis for

ℝ3?

No, because these vectors form a matrix with only 2 pivot columns. Therefore, these vectors form a basis for a two-dimensional subspace of

ℝ3.

No, because these two vectors are linearly dependent.

Yes, because these two vectors are linearly independent.

Yes, because these vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.20

Determine if the set is a basis for

ℝ3.

Justify your answer.

1

5

−2

,

2

−3

−3

,

3

−3

11

,

0

7

−5

Is the given set a basis for

ℝ3?

No, because these vectors form the columns of a

3×3

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these vectors do not form the columns of a

3×3

matrix. A set that contains more vectors than there are entries is linearly dependent.

Yes, because these vectors form the columns of an invertible

3×3

matrix. A set that contains more vectors than there are entries is linearly independent.

Yes, because these vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

2.8.20

Determine if the set is a basis for

ℝ3.

Justify your answer.

1

5

−2

,

7

−9

5

,

7

−7

−11

,

0

3

−7

Is the given set a basis for

ℝ3?

Yes, because these vectors form the columns of an invertible

3×3

matrix. By the invertible matrix theorem, the following statements are equivalent: A is an invertible matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

No, because these vectors do not form the columns of a

3×3

matrix. A set that contains more vectors than there are entries is linearly dependent.

No, because these vectors form the columns of a

3×3

matrix that is not invertible. By the invertible matrix theorem, the following statements are equivalent: A is a singular matrix, the columns of A form a linearly independent set, and the columns of A span

ℝn.

Yes, because these vectors form the columns of an invertible

3×3

matrix. A set that contains more vectors than there are entries is linearly independent.

2.8.21

Determine whether the statement below is true or false. Justify the answer.

A subspace of

ℝn

is any set H such that (i) the zero vector is in H, (ii)

u,

v,

and

u+v

are in H, and (iii) c is a scalar and

cu

is in H.

Choose the correct answer below.

The statement is false. It must also be separately specified that

u−v

are in H when

u

and

v

are in H.

The statement is true. Any set of elements of

ℝn

that satisfies conditions (i), (ii), and (iii) is a subspace by definition.

The statement is false. The zero vector does not need to be in a set for the set to be a subspace.

The statement is false. Conditions (ii) and (iii) must be satisfied for each

u

and

v

in H, which is not specified in the given statement.

The statement is false. It must also be separately specified that

au+bv

are in H when

u

and

v

are in H and a and b are scalars.

2.8.22

Determine whether the statement below is true or false. Justify the answer.

A subset H of

ℝn

is a subspace if the zero vector is in H.

Choose the correct answer below.

This statement is false. For each u and v in H, the product uv must also be in H.

This statement is false. The subset H is a subspace if the zero vector is not in H.

This statement is false. For each u and v in H and each scalar c, the sum

u+v

and the vector cu must also be in H.

This statement is true. This is the definition of a subspace.

2.8.23

Determine whether the statement below is true or false. Justify the answer.

If

v1, ..., vp

are in

ℝn,

then

S=Spanv1, ..., vp

is the same as the column space of the matrix

A=v1   •••  vp.

Choose the correct answer below.

The statement is true. The vectors

v1, ..., vp

always form a basis of both S and the column space of A.

The statement is true. The column space of A and S are both the set of the vectors

v1, ..., vp

and the zero vector.

The statement is true. The column space of A and S are both the set of all linear combinations of

v1, ..., vp.

The statement is false. There are infinitely many vectors in S but finitely many vectors in the column space of A.

The statement is false. There are finitely many vectors in S but infinitely many vectors in the column space of A.

The statement is false. S is the same as the column space of A only if the columns of A are linearly independent.

2.8.23

Determine whether the statement below is true or false. Justify the answer.

If

v1, ..., vp

are in

ℝn,

then

S=Spanv1, ..., vp

is the same as the column space of the matrix

A=v1   •••  vp.

Choose the correct answer below.

The statement is true. The column space of A and S are both the set of all linear combinations of

v1, ..., vp.

The statement is true. The vectors

v1, ..., vp

always form a basis of both S and the column space of A.

The statement is true. The column space of A and S are both the set of the vectors

v1, ..., vp

and the zero vector.

The statement is false. S is the same as the column space of A only if the columns of A are linearly independent.

The statement is false. There are infinitely many vectors in S but finitely many vectors in the column space of A.

The statement is false. There are finitely many vectors in S but infinitely many vectors in the column space of A.

2.8.24

Determine whether the statement below is true or false. Justify the answer.

Given vectors

v1,

...,

vp

in

ℝn,

the set of all linear combinations of these vectors is a subspace of

ℝn.

Choose the correct answer below.

This statement is false. This set is a subspace of

ℝp.

This statement is true. This set satisfies all properties of a subspace.

This statement is false. This set does not contain the zero vector.

This statement is false. This set is a subspace of

ℝn+p.

2.8.24

Determine whether the statement below is true or false. Justify the answer.

Given vectors

v1,

...,

vp

in

ℝn,

the set of all linear combinations of these vectors is a subspace of

ℝn.

Choose the correct answer below.

This statement is false. This set does not contain the zero vector.

This statement is false. This set is a subspace of

ℝp.

This statement is false. This set is a subspace of

ℝn+p.

This statement is true. This set satisfies all properties of a subspace.

2.8.24

Determine whether the statement below is true or false. Justify the answer.

Given vectors

v1,

...,

vp

in

ℝn,

the set of all linear combinations of these vectors is a subspace of

ℝn.

Choose the correct answer below.

This statement is false. This set is a subspace of

ℝn+p.

This statement is false. This set does not contain the zero vector.

This statement is false. This set is a subspace of

ℝp.

This statement is true. This set satisfies all properties of a subspace.

2.8.25

Determine whether the statement below is true or false. Justify the answer.

The set of all solutions of a system of m homogeneous equations in n unknowns is a subspace of

ℝm.

Choose the correct answer below.

The statement is true. The described set is the null space of an

m×n

matrix A. This set is a subspace of

ℝm.

The statement is false. The described set is the column space of an

m×n

matrix A. This set is a subspace of

ℝn.

The statement is false. The described set is the null space of an

m×n

matrix A. This set is a subspace of

ℝn.

The statement is false. The described set is only a subspace of

ℝm

if

m<n.

The statement is true. The described set is the column space of an

m×n

matrix A. This set is a subspace of

ℝm.

2.8.25

Determine whether the statement below is true or false. Justify the answer.

The set of all solutions of a system of m homogeneous equations in n unknowns is a subspace of

ℝm.

Choose the correct answer below.

The statement is false. The described set is only a subspace of

ℝm

if

m<n.

The statement is false. The described set is the null space of an

m×n

matrix A. This set is a subspace of

ℝn.

The statement is true. The described set is the null space of an

m×n

matrix A. This set is a subspace of

ℝm.

The statement is false. The described set is the column space of an

m×n

matrix A. This set is a subspace of

ℝn.

The statement is true. The described set is the column space of an

m×n

matrix A. This set is a subspace of

ℝm.

2.8.26

Determine whether the statement below is true or false. Justify the answer.

The null space of an

m×n

matrix is a subspace of

ℝn.

Choose the correct answer below.

This statement is false. This set is not closed under scalar multiplication.

This statement is true. For an

m×n

matrix A, the solutions of

Ax=0

are vectors in

ℝn

and satisfy the properties of a vector space.

This statement is false. For an

m×n

matrix A, the solutions of

Ax=0

belong to

ℝm.

This statement is false. The null space of a matrix does not contain the zero vector.

2.8.27

Determine whether the statement below is true or false. Justify the answer.

The columns of an invertible

n×n

matrix form a basis for

ℝn.

Choose the correct answer below.

The statement is false. There are not enough columns in an

n×n

matrix to form a basis for

ℝn.

The statement is true. The columns of an invertible

n×n

matrix are linearly independent and span

ℝn,

so they form a basis for

ℝn.

The statement is false. The columns of an invertible

n×n

matrix are not linearly independent. This means they cannot form a basis for any vector space, including

ℝn.

The statement is false. The columns of an invertible

n×n

matrix are linearly independent, but they do not span

ℝn.

This means they cannot form a basis for

ℝn.

The statement is true. There are n columns in an

n×n

matrix, each of which is a vector in

ℝn.

Any set of n vectors in

ℝn

must form a basis for

ℝn.

2.8.27

Determine whether the statement below is true or false. Justify the answer.

The columns of an invertible

n×n

matrix form a basis for

ℝn.

Choose the correct answer below.

The statement is true. The columns of an invertible

n×n

matrix are linearly independent and span

ℝn,

so they form a basis for

ℝn.

The statement is false. The columns of an invertible

n×n

matrix are not linearly independent. This means they cannot form a basis for any vector space, including

ℝn.

The statement is true. There are n columns in an

n×n

matrix, each of which is a vector in

ℝn.

Any set of n vectors in

ℝn

must form a basis for

ℝn.

The statement is false. There are not enough columns in an

n×n

matrix to form a basis for

ℝn.

The statement is false. The columns of an invertible

n×n

matrix are linearly independent, but they do not span

ℝn.

This means they cannot form a basis for

ℝn.

2.8.31

A matrix A and an echelon form of A are shown below. Find a basis for Col A and a basis for Nul A.

A

=

5 8 17 −39 5 2 −7 9 2 1 −2 2

~

1 2 5 −11 0 1 4 −8 0 0 0 0

Find a basis for Col A.

	5
	5
	2

,

	8
	2
	1

(Simplify your answer. Use a comma to separate answers as needed.)

Find a basis for Nul A.

	3
	−4
	1
	0

,

	−5
	8
	0
	1

(Simplify your answer. Use a comma to separate answers as needed.)

2.8.31

A matrix A and an echelon form of A are shown below. Find a basis for Col A and a basis for Nul A.

A

=

3 6 21 −15 7 15 54 −40 3 8 31 −25

~

1 3 12 −10 0 1 5 −5 0 0 0 0

Find a basis for Col A.

	3
	7
	3

,

	6
	15
	8

(Simplify your answer. Use a comma to separate answers as needed.)

Find a basis for Nul A.

	3
	−5
	1
	0

,

	−5
	5
	0
	1

(Simplify your answer. Use a comma to separate answers as needed.)

2.8.31

A matrix A and an echelon form of A are shown below. Find a basis for Col A and a basis for Nul A.

A

=

3 7 19 −38 6 10 22 −44 3 7 19 −38

~

1 3 9 −18 0 1 4 −8 0 0 0 0

Find a basis for Col A.

	3
	6
	3

,

	7
	10
	7

(Simplify your answer. Use a comma to separate answers as needed.)

Find a basis for Nul A.

	3
	−4
	1
	0

,

	−6
	8
	0
	1

(Simplify your answer. Use a comma to separate answers as needed.)

2.8.33

A matrix A and an echelon form of A are shown below. Find a basis for Col A and a basis for Nul A.

A

=

1 15 25 −1 1 −1 15 29 4 −7 −2 −5 −5 6 −4 3 15 21 −6 9

~

1 10 16 0 5 0 5 9 0 −2 0 0 0 1 2 0 0 0 0 0

Find a basis for Col A.

	1
	−1
	−2
	3

,

	15
	15
	−5
	15

,

	−1
	4
	6
	−6

(Use a comma to separate answers as needed. Type an integer or simplified fraction for each matrix element.)

Find a basis for Nul A.

	2
	−95
	1
	0
	0

,

	−9
	25
	0
	−2
	1

(Use a comma to separate answers as needed. Type an integer or simplified fraction for each matrix element.)

2.9.3

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

2

−3

,

b2=

−1

4

,

x=

6

−4

[x]B=

	4
	2

2.9.3

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

2

−3

,

b2=

−1

3

,

x=

6

−6

[x]B=

	4
	2

2.9.3

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

2

−5

,

b2=

−1

4

,

x=

4

−7

[x]B=

	3
	2

2.9.5

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

1

2

−4

,

b2=

−4

−7

15

,

x=

−7

−12

26

[x]B=

	1
	2

2.9.5

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

1

2

−3

,

b2=

−4

−7

11

,

x=

−11

−18

29

[x]B=

	5
	4

2.9.5

The vector x is in a subspace H with a basis

B={b1,b2}.

Find the B-coordinate vector of

x.

b1=

1

4

−3

,

b2=

−3

−11

8

,

x=

−5

−17

12

[x]B=

	4
	3

2.9.9

Given below is a matrix A and an echelon form of A. Find bases for

Col A

and

Nul A,

and then state the dimensions of these subspaces.

A=

	1	3	5	−4
	5	15	1	3
	3	9	−6	7
	4	12	0	13

~

	1	3	3	6
	0	0	7	−8
	0	0	0	7
	0	0	0	0

A basis for

Col A

is given by

	1
	5
	3
	4

,

	5
	1
	−6
	0

,

	−4
	3
	7
	13

.

(Use a comma to separate answers as needed.)

The dimension of

Col A

is

3.

(Type an integer.)

A basis for

Nul A

is given by

	−3
	1
	0
	0

.

(Use a comma to separate answers as needed.)

The dimension of

Nul A

is

1.

(Type an integer.)

2.9.9

Given below is a matrix A and an echelon form of A. Find bases for

Col A

and

Nul A,

and then state the dimensions of these subspaces.

A=

	1	2	5	−6
	3	6	1	4
	6	12	−5	10
	4	8	0	12

~

	1	2	6	3
	0	0	3	−8
	0	0	0	3
	0	0	0	0

A basis for

Col A

is given by

	1
	3
	6
	4

,

	5
	1
	−5
	0

,

	−6
	4
	10
	12

.

(Use a comma to separate answers as needed.)

The dimension of

Col A

is

3.

(Type an integer.)

A basis for

Nul A

is given by

	−2
	1
	0
	0

.

(Use a comma to separate answers as needed.)

The dimension of

Nul A

is

1.

(Type an integer.)

2.9.9

Given below is a matrix A and an echelon form of A. Find bases for

Col A

and

Nul A,

and then state the dimensions of these subspaces.

A=

	1	2	3	−7
	4	8	1	5
	3	6	−6	12
	6	12	0	19

~

	1	2	3	4
	0	0	4	−9
	0	0	0	4
	0	0	0	0

A basis for

Col A

is given by

	1
	4
	3
	6

,

	3
	1
	−6
	0

,

	−7
	5
	12
	19

.

(Use a comma to separate answers as needed.)

The dimension of

Col A

is

3.

(Type an integer.)

A basis for

Nul A

is given by

	−2
	1
	0
	0

.

(Use a comma to separate answers as needed.)

The dimension of

Nul A

is

1.

(Type an integer.)

2.9.11

Find the bases for Col A and Nul A, and then state the dimension of these subspaces for the matrix A and an echelon form of A below.

A=

	1	2	−1	−1	−7
	2	5	0	1	−6
	−3	−9	−3	−1	17
	3	10	5	4	−9

~

	1	2	−1	−1	−7
	0	1	2	3	8
	0	0	0	1	4
	0	0	0	0	0

A basis for Col A is given by

	1
	2
	−3
	3

,

	2
	5
	−9
	10

,

	−1
	1
	−1
	4

.

(Use a comma to separate vectors as needed.)

The dimension of Col A is

3.

A basis for Nul A is given by

	5
	−2
	1
	0
	0

,

	−5
	4
	0
	−4
	1

.

(Use a comma to separate vectors as needed.)

The dimension of Nul A is

2.

2.9.11

Find the bases for Col A and Nul A, and then state the dimension of these subspaces for the matrix A and an echelon form of A below.

A=

	1	2	3	0	−4
	2	5	10	3	0
	−3	−9	−21	−4	8
	3	10	25	7	0

~

	1	2	3	0	−4
	0	1	4	3	8
	0	0	0	1	4
	0	0	0	0	0

A basis for Col A is given by

	1
	2
	−3
	3

,

	2
	5
	−9
	10

,

	0
	3
	−4
	7

.

(Use a comma to separate vectors as needed.)

The dimension of Col A is

3.

A basis for Nul A is given by

	5
	−4
	1
	0
	0

,

	−4
	4
	0
	−4
	1

.

(Use a comma to separate vectors as needed.)

The dimension of Nul A is

2.

2.9.11

Find the bases for Col A and Nul A, and then state the dimension of these subspaces for the matrix A and an echelon form of A below.

A=

	1	3	3	2	0
	2	7	10	6	3
	−3	−12	−21	−7	6
	3	13	25	9	−3

~

	1	3	3	2	0
	0	1	4	2	3
	0	0	0	1	3
	0	0	0	0	0

A basis for Col A is given by

	1
	2
	−3
	3

,

	3
	7
	−12
	13

,

	2
	6
	−7
	9

.

(Use a comma to separate vectors as needed.)

The dimension of Col A is

3.

A basis for Nul A is given by

	9
	−4
	1
	0
	0

,

	−3
	3
	0
	−3
	1

.

(Use a comma to separate vectors as needed.)

The dimension of Nul A is

2.

2.9.12

Find the bases for Col A and Nul A, and then state the dimension of these subspaces for the matrix A and an echelon form of A below.

A=

	1	2	−3	6	4
	5	10	−4	−3	13
	4	8	−5	3	8
	−2	−4	3	−3	−8

~

	1	2	−3	6	4
	0	0	1	−3	0
	0	0	0	0	−8
	0	0	0	0	0

A basis for Col A is given by

	1
	5
	4
	−2

,

	−3
	−4
	−5
	3

,

	4
	13
	8
	−8

.

(Use a comma to separate vectors as needed.)

The dimension of Col A is

3.

A basis for Nul A is given by

	−2
	1
	0
	0
	0

,

	3
	0
	3
	1
	0

.

(Use a comma to separate vectors as needed.)

The dimension of Nul A is

2.

2.9.13

Find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace?

	1
	−3
	4
	−6

,

	−3
	9
	−12
	18

,

	2
	−1
	5
	6

,

	−5
	7
	−2
	12

A basis for the subspace is given by

	1
	−3
	4
	−6

,

	2
	−1
	5
	6

,

	−5
	7
	−2
	12

.

(Use a comma to separate answers as needed.)

The dimension of this subspace is

3.

(Type an integer.)

2.9.13

Find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace?

	1
	−5
	6
	−2

,

	−3
	15
	−18
	6

,

	2
	−1
	6
	3

,

	−5
	3
	−3
	9

A basis for the subspace is given by

	1
	−5
	6
	−2

,

	2
	−1
	6
	3

,

	−5
	3
	−3
	9

.

(Use a comma to separate answers as needed.)

The dimension of this subspace is

3.

(Type an integer.)

2.9.13

Find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace?

	1
	−6
	3
	−2

,

	−2
	12
	−6
	4

,

	3
	−1
	4
	5

,

	−8
	5
	−6
	10

A basis for the subspace is given by

	1
	−6
	3
	−2

,

	3
	−1
	4
	5

,

	−8
	5
	−6
	10

.

(Use a comma to separate answers as needed.)

The dimension of this subspace is

3.

(Type an integer.)

2.9.15

Suppose a

3×5

matrix A has

three

pivot columns. Is Col

A=ℝ3?

Is Nul

A=ℝ2?

Explain your answers.

Is Col

A=ℝ3?

Explain your answer. Choose the correct answer and reasoning below.

No, because a

3×5

matrix exists in

ℝ5.

If its pivot columns form a

3-dimensional

basis, then Col A is isomorphic to

ℝ3

but is not strictly equal to

ℝ3.

Yes, because there are

three

pivot columns in A. These columns form a basis in

three

dimensions. Any

3-dimensional

basis spans

ℝ3.

Yes, because the column space of a

3×5

matrix is a subspace of

ℝ3.

There is a pivot in each row, so the column space is

3-dimensional.

Since any

3-dimensional

subspace of

ℝ3

is

ℝ3,

Col

A=ℝ3.

No, Col

A=ℝ2.

The number of pivot columns is equal to the dimension of the null space. Since the sum of the dimensions of the null space and column space equals the number of columns in the matrix, the dimension of the column space must be

2.

Since any

2-dimensional

basis is equal to

ℝ2,

Col

A=ℝ2.

Is Nul

A=ℝ2?

Explain your answer. Choose the correct answer and reasoning below.

Yes, because the linearly dependent vectors in A form a basis in

two

dimensions. Any basis in

two

dimensions is also a basis for

ℝ2.

Therefore, Nul

A=ℝ2.

Yes, because a

3×5

matrix exists in

ℝ3.

Therefore, if its null space is

2-dimensional

and contained within

ℝ3,

it must be equal to

ℝ2.

No, because the null space of a

3×5

matrix is a subspace of

ℝ5.

Although

dim Nul A=2,

it is not strictly equal to

ℝ2

because each vector in Nul A has

five

components. Each vector in

ℝ2

has

two

components. Therefore, Nul A is isomorphic to

ℝ2,

but not equal.

No, because although the null space is

2-dimensional,

its basis consists of

three

vectors and not

two.

Therefore, it cannot be equal to

ℝ2.

2.9.15

Suppose a

3×6

matrix A has

three

pivot columns. Is Col

A=ℝ3?

Is Nul

A=ℝ3?

Explain your answers.

Is Col

A=ℝ3?

Explain your answer. Choose the correct answer and reasoning below.

No, Col

A=ℝ3.

The number of pivot columns is equal to the dimension of the null space. Since the sum of the dimensions of the null space and column space equals the number of columns in the matrix, the dimension of the column space must be

3.

Since any

3-dimensional

basis is equal to

ℝ3,

Col

A=ℝ3.

No, because a

3×6

matrix exists in

ℝ6.

If its pivot columns form a

3-dimensional

basis, then Col A is isomorphic to

ℝ3

but is not strictly equal to

ℝ3.

Yes, because there are

three

pivot columns in A. These columns form a basis in

three

dimensions. Any

3-dimensional

basis spans

ℝ3.

Yes, because the column space of a

3×6

matrix is a subspace of

ℝ3.

There is a pivot in each row, so the column space is

3-dimensional.

Since any

3-dimensional

subspace of

ℝ3

is

ℝ3,

Col

A=ℝ3.

Is Nul

A=ℝ3?

Explain your answer. Choose the correct answer and reasoning below.

No, because although the null space is

3-dimensional,

its basis consists of

three

vectors and not

three.

Therefore, it cannot be equal to

ℝ3.

Yes, because a

3×6

matrix exists in

ℝ3.

Therefore, if its null space is

3-dimensional

and contained within

ℝ3,

it must be equal to

ℝ3.

Yes, because the linearly dependent vectors in A form a basis in

three

dimensions. Any basis in

three

dimensions is also a basis for

ℝ3.

Therefore, Nul

A=ℝ3.

No, because the null space of a

3×6

matrix is a subspace of

ℝ6.

Although

dim Nul A=3,

it is not strictly equal to

ℝ3

because each vector in Nul A has

six

components. Each vector in

ℝ3

has

three

components. Therefore, Nul A is isomorphic to

ℝ3,

but not equal.

2.9.17

Determine whether the statement below is true or false. Justify the answer.

If

B=v1,...,vp

is a basis for a subspace H and if

x=c1v1+...+cpvp,

then

c1,...,cp

are the coordinates of x relative to the basis

B.

Choose the correct answer below.

The statement is false because the coordinate vector

[x]B

is composed of the coordinates

c1,...,cp

only if the vector x in

ℝp

is equal to

c1v1+...+cpvp.

The statement is false because x is

v1

coordinates in the direction of

c1,

v2

coordinates in the direction of

c2,

and so on.

The statement is true because the coordinates

c1,...,cp

are the same as the coordinates of x relative to the xy-plane.

The statement is true because any coordinate in a subspace H, with basis

B,

can only be written in one way as a linear combination of basis vectors. The linear combination gives a unique coordinate vector

[x]B

that is composed of the coordinates of x relative to

B.

2.9.17

Determine whether the statement below is true or false. Justify the answer.

If

B=v1,...,vp

is a basis for a subspace H and if

x=c1v1+...+cpvp,

then

c1,...,cp

are the coordinates of x relative to the basis

B.

Choose the correct answer below.

The statement is false because x is

v1

coordinates in the direction of

c1,

v2

coordinates in the direction of

c2,

and so on.

The statement is true because any coordinate in a subspace H, with basis

B,

can only be written in one way as a linear combination of basis vectors. The linear combination gives a unique coordinate vector

[x]B

that is composed of the coordinates of x relative to

B.

The statement is false because the coordinate vector

[x]B

is composed of the coordinates

c1,...,cp

only if the vector x in

ℝp

is equal to

c1v1+...+cpvp.

The statement is true because the coordinates

c1,...,cp

are the same as the coordinates of x relative to the xy-plane.

2.9.17

Determine whether the statement below is true or false. Justify the answer.

If

B=v1,...,vp

is a basis for a subspace H and if

x=c1v1+...+cpvp,

then

c1,...,cp

are the coordinates of x relative to the basis

B.

Choose the correct answer below.

The statement is false because the coordinate vector

[x]B

is composed of the coordinates

c1,...,cp

only if the vector x in

ℝp

is equal to

c1v1+...+cpvp.

The statement is false because x is

v1

coordinates in the direction of

c1,

v2

coordinates in the direction of

c2,

and so on.

The statement is true because the coordinates

c1,...,cp

are the same as the coordinates of x relative to the xy-plane.

The statement is true because any coordinate in a subspace H, with basis

B,

can only be written in one way as a linear combination of basis vectors. The linear combination gives a unique coordinate vector

[x]B

that is composed of the coordinates of x relative to

B.

2.9.18

Determine whether the statement below is true or false. Justify the answer.

If

B

is a basis for a subspace H, then each vector in H can be written in only one way as a linear combination of the vectors in

B.

Choose the correct answer below.

The statement is true. Suppose

B=v1,...,vp

and x is a vector in H that can be generated two ways. Say,

x=c1v1+...+cpvp

and

x=d1v1+...+dpvp,

then

0=x−x=c1−d1v1+...+cp−dpvp.

Therefore,

cp=dp

and x can only be generated in one way.

The statement is true. All bases for a subspace H are linearly independent and therefore each vector in H can only be generated as one unique linear combination of the vectors in

B.

The statement is false. Suppose

B=v1,...,vp

and x is a vector in H. The vector x can be generated in a multiple of ways based on the values of the vectors in the set

B=v1,...,vp.

The statement is false. Bases for a subspace H may be linear dependent and therefore there can be multiple solutions for the same vector x in H.

2.9.18

Determine whether the statement below is true or false. Justify the answer.

If

B

is a basis for a subspace H, then each vector in H can be written in only one way as a linear combination of the vectors in

B.

Choose the correct answer below.

The statement is true. All bases for a subspace H are linearly independent and therefore each vector in H can only be generated as one unique linear combination of the vectors in

B.

The statement is true. Suppose

B=v1,...,vp

and x is a vector in H that can be generated two ways. Say,

x=c1v1+...+cpvp

and

x=d1v1+...+dpvp,

then

0=x−x=c1−d1v1+...+cp−dpvp.

Therefore,

cp=dp

and x can only be generated in one way.

The statement is false. Bases for a subspace H may be linear dependent and therefore there can be multiple solutions for the same vector x in H.

The statement is false. Suppose

B=v1,...,vp

and x is a vector in H. The vector x can be generated in a multiple of ways based on the values of the vectors in the set

B=v1,...,vp.

2.9.19

Determine whether the statement below is true or false. Justify the answer.

Each line in

ℝn

is a one-dimensional subspace of

ℝn.

Choose the correct answer below.

The statement is false. Any subspace of

ℝn

must be at least n-dimensional.

The statement is false. Any subspace of

ℝn

must contain the zero-vector. Therefore, a line can only be a one-dimensional subspace of

ℝn

if it passes through the origin.

The statement is true. Any line in

ℝn

satisfies all three requirements of a subspace.

The statement is true. Any one-dimensional subspace of

ℝn

must be a line.

2.9.19

Determine whether the statement below is true or false. Justify the answer.

Each line in

ℝn

is a one-dimensional subspace of

ℝn.

Choose the correct answer below.

The statement is true. Any one-dimensional subspace of

ℝn

must be a line.

The statement is false. Any subspace of

ℝn

must contain the zero-vector. Therefore, a line can only be a one-dimensional subspace of

ℝn

if it passes through the origin.

The statement is true. Any line in

ℝn

satisfies all three requirements of a subspace.

The statement is false. Any subspace of

ℝn

must be at least n-dimensional.

2.9.21

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimension of Col A is the number of pivot columns in A.

Choose the correct answer below.

The statement is false. The dimension of Col A cannot be determined without the size of matrix A.

The statement is false. The number of pivot columns determines the dimension of the null space, not the column space.

The statement is true. The pivot columns of A form a basis for Col A. Therefore, the number of pivot columns of A is the same as the dimension of Col A.

The statement is true. The number of pivot columns is equal to the number of free variables in the equation

Ax=0.

The number of free variables in

Ax=0

is equal to the dimension of the column space.

2.9.21

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimension of Col A is the number of pivot columns in A.

Choose the correct answer below.

The statement is false. The number of pivot columns determines the dimension of the null space, not the column space.

The statement is true. The pivot columns of A form a basis for Col A. Therefore, the number of pivot columns of A is the same as the dimension of Col A.

The statement is false. The dimension of Col A cannot be determined without the size of matrix A.

The statement is true. The number of pivot columns is equal to the number of free variables in the equation

Ax=0.

The number of free variables in

Ax=0

is equal to the dimension of the column space.

2.9.22

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimension of Nul A is the number of variables in the equation

Ax=0.

Choose the correct answer below.

The statement is true. The dimension of Nul A is the same as the amount of vectors in the set

x=x1,...,xp

that satisfy the equation

Ax=0.

The statement is true. The number of total variables involved in solving the equation

Ax=0

is the dimension of Nul A.

The statement is false. The dimension of Nul A is the number of free variables in the equation

Ax=0.

The statement is false. The dimension of Nul A is the number of variables in the equation

Ax=0

minus the number of free variables in the equation

Ax=0.

2.9.22

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimension of Nul A is the number of variables in the equation

Ax=0.

Choose the correct answer below.

The statement is true. The dimension of Nul A is the same as the amount of vectors in the set

x=x1,...,xp

that satisfy the equation

Ax=0.

The statement is false. The dimension of Nul A is the number of free variables in the equation

Ax=0.

The statement is true. The number of total variables involved in solving the equation

Ax=0

is the dimension of Nul A.

The statement is false. The dimension of Nul A is the number of variables in the equation

Ax=0

minus the number of free variables in the equation

Ax=0.

2.9.23

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimensions of Col A and Nul A add up to the total number of columns in A.

Choose the correct answer below.

The statement is true. The Rank Theorem states that if matrix A has n columns, then

rank A+dim Nul A=n.

Since rank A is the same as

dim Col A,

the

dimensions of

Col A

and

Nul A

add up to the total number of columns in A.

The statement is true. Col A and Nul A are both subspaces of

ℝn,

where n is the number of columns in matrix A.

The statement is false. The column space and null space sometimes intersect. Therefore, the sum of

dim Col A

and

dim Nul A

is the number of columns of A, only if the

Col A

and

Nul A

are disjoint.

The statement is false. The sum of

dim Col A

and

dim Nul A

is the number of rows of A.

2.9.23

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimensions of Col A and Nul A add up to the total number of columns in A.

Choose the correct answer below.

The statement is true. The Rank Theorem states that if matrix A has n columns, then

rank A+dim Nul A=n.

Since rank A is the same as

dim Col A,

the

dimensions of

Col A

and

Nul A

add up to the total number of columns in A.

The statement is false. The column space and null space sometimes intersect. Therefore, the sum of

dim Col A

and

dim Nul A

is the number of columns of A, only if the

Col A

and

Nul A

are disjoint.

The statement is true. Col A and Nul A are both subspaces of

ℝn,

where n is the number of columns in matrix A.

The statement is false. The sum of

dim Col A

and

dim Nul A

is the number of rows of A.

2.9.24

Determine whether the statement below is true or false. Justify the answer. Here, A is an

m×n

matrix.

The dimension of the column space of A is rank A.

Choose the correct answer below.

The statement is true. The rank of matrix A is the dimension of the column space of A.

The statement is true. The rank of matrix A is equal to n, which is also equal to the dimension of the column space of A.

The statement is false. The rank of matrix A is equal to the dimension of the column space of A plus the dimension of the null space of A.

The statement is false. The rank of matrix A is equal to the number of columns plus the number of rows of A, or rank

A=m+n.

2.9.25

Determine whether the statement below is true or false. Justify the answer.

If a set of p vectors spans a p-dimensional subspace H of

ℝn,

then these vectors form a basis of H.

Choose the correct answer below.

The statement is false. Although the set of vectors spans H, there is not enough information to conclude that they form a basis of H.

The statement is false. Only vectors that span H and are linearly independent will form a basis of H. Since the set contains too many vectors, the spanning set cannot possibly be linearly independent.

The statement is true. If a set of p vectors spans a p-dimensional subspace H of

ℝn,

then these vectors must be linearly independent. Any linearly independent spanning set of p vectors forms a basis in p dimensions.

The statement is true. Any spanning set in H will form a basis of H.

2.9.27

If the subspace of all solutions of

Ax=0

has a basis consisting of

three

vectors and if A is a

5×7

matrix, what is the rank of A?

rank

A=4

(Type a whole number.)

2.9.29

If the rank of a

6×8

matrix A is

2,

what is the dimension of the solution space

Ax=0?

The dimension of the solution space is

6.

2.9.29

If the rank of a

5×8

matrix A is

2,

what is the dimension of the solution space

Ax=0?

The dimension of the solution space is

6.

2.9.37

Let

H=Span

v1,v2

and

B=v1,v2.

Show that x is in H, and find the

B-coordinate

vector of

x,

when

v1,

v2,

and x are as below.

v1=

	12
	−5
	11
	9

,

v2=

	15
	−8
	14
	12

,

x=

	20
	−13
	19
	17

Reduce the augmented matrix

	v1	v2	x

to reduced echelon form.

	12	15	20
	−5	−8	−13
	11	14	19
	9	12	17

~

	1	0	−53
	0	1	83
	0	0	0
	0	0	0

How can it be shown that x is in H?

The augmented matrix is upper triangular and row equivalent to

	B	x

,

therefore x is in H because H is the

Spanv1,v2

and

B=v1,v2.

The augmented matrix shows that the system of equations is consistent and therefore x is in H.

The last two rows of the augmented matrix has zero for all entries and this implies that x must be in H.

The first two columns of the augmented matrix are pivot columns and therefore x is in H.

This implies that the

B-coordinate

vector is

	x

B=

	−53
	83

.

2.9.37

Let

H=Span

v1,v2

and

B=v1,v2.

Show that x is in H, and find the

B-coordinate

vector of

x,

when

v1,

v2,

and x are as below.

v1=

	13
	−5
	12
	7

,

v2=

	16
	−8
	15
	10

,

x=

	21
	−13
	20
	15

Reduce the augmented matrix

	v1	v2	x

to reduced echelon form.

	13	16	21
	−5	−8	−13
	12	15	20
	7	10	15

~

	1	0	−53
	0	1	83
	0	0	0
	0	0	0

How can it be shown that x is in H?

The augmented matrix is upper triangular and row equivalent to

	B	x

,

therefore x is in H because H is the

Spanv1,v2

and

B=v1,v2.

The last two rows of the augmented matrix has zero for all entries and this implies that x must be in H.

The first two columns of the augmented matrix are pivot columns and therefore x is in H.

The augmented matrix shows that the system of equations is consistent and therefore x is in H.

This implies that the

B-coordinate

vector is

	x

B=

	−53
	83

.

2.9.37

Let

H=Span

v1,v2

and

B=v1,v2.

Show that x is in H, and find the

B-coordinate

vector of

x,

when

v1,

v2,

and x are as below.

v1=

	10
	−7
	9
	8

,

v2=

	13
	−10
	12
	11

,

x=

	17
	−14
	16
	15

Reduce the augmented matrix

	v1	v2	x

to reduced echelon form.

	10	13	17
	−7	−10	−14
	9	12	16
	8	11	15

~

	1	0	−43
	0	1	73
	0	0	0
	0	0	0

How can it be shown that x is in H?

The first two columns of the augmented matrix are pivot columns and therefore x is in H.

The augmented matrix shows that the system of equations is consistent and therefore x is in H.

The augmented matrix is upper triangular and row equivalent to

	B	x

,

therefore x is in H because H is the

Spanv1,v2

and

B=v1,v2.

The last two rows of the augmented matrix has zero for all entries and this implies that x must be in H.

This implies that the

B-coordinate

vector is

	x

B=

	−43
	73

.

2.9.38

Let

H=Spanv1, v2, v3

and

B=v1, v2, v3.

Show that

B

is a basis for H and x is in H, and find the

B-coordinate

vector of x for the given vectors.

v1=

	−7
	2
	−9
	6

,

v2=

	8
	−3
	6
	−3

,

v3=

	−9
	4
	−7
	−3

,

x=

	−27
	7
	−43
	24

Reduce the augmented matrix

	v1	v2	v3	x

to reduced echelon form.

	−7	8	−9	−27
	2	−3	4	7
	−9	6	−7	−43
	6	−3	−3	24

~

	1	0	0	6
	0	1	0	3
	0	0	1	1
	0	0	0	0

How can it be shown that

B

is a basis for H?

H is the

Spanv1, v2, v3

and

B=v1, v2, v3

so therefore

B

must form a basis for H.

The first three columns of the augmented matrix are pivot columns and therefore

B

forms a basis for H.

The augmented matrix shows that the system of equations is consistent and therefore

B

forms a basis for H.

The augmented matrix is upper triangular and row equivalent to

	B	x

therefore,

B

forms a basis for H.

How can it be shown that x is in H?

The first three columns of the augmented matrix are pivot columns and therefore x is in H.

The augmented matrix is upper triangular and row equivalent to

	B	x

,

therefore x is in H because H is the

Spanv1, v2, v3

and

B=v1, v2, v3.

The augmented matrix shows that the system of equations is consistent and therefore x is in H.

The last row of the augmented matrix has zero for all entries and this implies that x must be in H.

The

B-coordinate

vector of is

	x

B

=

	6
	3
	1

.

3.1.1

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.

3	0	4
2	3	3
0	4	−1

Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

−(0)(−2)+(3)(−3)−(4)(1)=.

Using this expansion, the determinant is

(0)(−2)−(3)(−3)+(4)(1)=.

Using this expansion, the determinant is

−(3)(−15)+(0)(−2)−(4)(8)=.

Using this expansion, the determinant is

(3)(−15)−(0)(−2)+(4)(8)=−13.

Compute the determinant using a cofactor expansion down the second column. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

−(3)(−15)+(0)(−2)−(4)(8)=.

Using this expansion, the determinant is

(3)(−15)−(0)(−2)+(4)(8)=.

Using this expansion, the determinant is

(0)(−2)−(3)(−3)+(4)(1)=.

Using this expansion, the determinant is

−(0)(−2)+(3)(−3)−(4)(1)=−13.

3.1.1

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.

3	0	4
2	4	2
0	4	−2

Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

−(0)(−4)+(4)(−6)−(4)(−2)=.

Using this expansion, the determinant is

(3)(−16)−(0)(−4)+(4)(8)=−16.

Using this expansion, the determinant is

−(3)(−16)+(0)(−4)−(4)(8)=.

Using this expansion, the determinant is

(0)(−4)−(4)(−6)+(4)(−2)=.

Compute the determinant using a cofactor expansion down the second column. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

(0)(−4)−(4)(−6)+(4)(−2)=.

Using this expansion, the determinant is

−(0)(−4)+(4)(−6)−(4)(−2)=−16.

Using this expansion, the determinant is

−(3)(−16)+(0)(−4)−(4)(8)=.

Using this expansion, the determinant is

(3)(−16)−(0)(−4)+(4)(8)=.

3.1.1

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.

2	0	4
3	4	3
0	5	−2

Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

(0)(−6)−(4)(−4)+(5)(−6)=.

Using this expansion, the determinant is

(2)(−23)−(0)(−6)+(4)(15)=14.

Using this expansion, the determinant is

−(2)(−23)+(0)(−6)−(4)(15)=.

Using this expansion, the determinant is

−(0)(−6)+(4)(−4)−(5)(−6)=.

Compute the determinant using a cofactor expansion down the second column. Select the correct choice below and fill in the answer box to complete your choice.

(Simplify your answer.)

Using this expansion, the determinant is

−(2)(−23)+(0)(−6)−(4)(15)=.

Using this expansion, the determinant is

−(0)(−6)+(4)(−4)−(5)(−6)=14.

Using this expansion, the determinant is

(0)(−6)−(4)(−4)+(5)(−6)=.

Using this expansion, the determinant is

(2)(−23)−(0)(−6)+(4)(15)=.

3.1.9

Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation.

9 0 0 5 8 7 3 −3 3 0 0 0 7 3 1 1

9	0	0	5
8	7	3	−3
3	0	0	0
7	3	1	1

=−30

(Simplify your answer.)

3.1.9

Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation.

5 0 0 5 6 7 3 −2 2 0 0 0 3 3 1 1

5	0	0	5
6	7	3	−2
2	0	0	0
3	3	1	1

=−20

(Simplify your answer.)

3.1.10

Compute the following determinant by cofactor expansions. At each step, choose the row or column that involves the least amount of computation.

3	−3	3	5
0	0	4	0
1	−6	−4	5
5	0	5	3

The determinant is

−120.

3.1.10

Compute the following determinant by cofactor expansions. At each step, choose the row or column that involves the least amount of computation.

4	−5	3	5
0	0	2	0
3	−2	−7	5
6	0	5	3

The determinant is

138.

3.1.15

The expansion of a 3×3 determinant can be remembered by this device. Write a second copy of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals. Add the downward diagonal products and subtract the upward products. Use this method to compute the following determinant. −1 −4 2 5 0 −1 0 −3 2

− − − + + + a11 a12 a13 a11 a12 a21 a22 a23 a21 a22 a31 a32 a33 a31 a32

−1	−4	2
5	0	−1
0	−3	2

=13

3.1.15

The expansion of a 3×3 determinant can be remembered by this device. Write a second copy of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals. Add the downward diagonal products and subtract the upward products. Use this method to compute the following determinant. −5 −4 −2 3 4 0 0 −4 −3

− − − + + + a11 a12 a13 a11 a12 a21 a22 a23 a21 a22 a31 a32 a33 a31 a32

−5	−4	−2
3	4	0
0	−4	−3

=48

3.1.20

State the row operation performed below and describe how it affects the determinant.

a	b
c	d

,

a	b
3c	3d

What row operation was performed?

The row operation

scales row 2 by one-third.

The row operation

adds 3 to row 2.

The row operation

subtracts 3 from row 2.

The row operation

scales row 2 by 3.

How does this affect the determinant?

The determinant is divided by

3.

The determinant is multiplied by 3.

The determinant is 0.

The determinant is unchanged.

3.1.20

State the row operation performed below and describe how it affects the determinant.

a	b
c	d

,

c	d
a	b

What row operation was performed?

The row operation

multiplies row 1 by d and row 2 by b.

The row operation

multiplies row 1 by c and row 2 by a.

The row operation

scales row 1 by c and row 2 by a.

The row operation

swaps rows 1 and 2.

How does this affect the determinant?

The determinant is 0.

The sign of the determinant is reversed.

The determinant is divided by

2.

The determinant is unchanged.

3.1.21

Explore the effects of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

8	4
2	5

,

8	4
2+8k	5+4k

What is the elementary row operation?

Replace row 2 with k times row 2.

Replace row 2 with k times row 1 plus row 2.

Replace row 2 with k times row 1.

Replace row 2 with row 1 plus k times row 2.

How does the row operation affect the determinant?

The determinant is decreased by

32k.

The determinant is increased by

64k.

The determinant is increased by

32k.

The determinant does not change.

3.1.21

Explore the effects of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

2	6
9	2

,

2	6
9+2k	2+6k

What is the elementary row operation?

Replace row 2 with k times row 1 plus row 2.

Replace row 2 with row 1 plus k times row 2.

Replace row 2 with k times row 2.

Replace row 2 with k times row 1.

How does the row operation affect the determinant?

The determinant is increased by

24k.

The determinant is increased by

12k.

The determinant is decreased by

12k.

The determinant does not change.

3.1.21

Explore the effects of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

3	6
6	2

,

3	6
6+3k	2+6k

What is the elementary row operation?

Replace row 2 with k times row 2.

Replace row 2 with k times row 1 plus row 2.

Replace row 2 with row 1 plus k times row 2.

Replace row 2 with k times row 1.

How does the row operation affect the determinant?

The determinant is decreased by

18k.

The determinant is increased by

18k.

The determinant is increased by

36k.

The determinant does not change.

3.1.22

Explore the effect of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

	a	b
	c	d

,

	a+kc	b+kd
	c	d

What is the elementary row operation?

Row 2 is replaced with the sum of itself and k times row 1.

Row 1 is replaced with the sum of itself and k times row 2.

Row 2 is multiplied by k.

Rows 1 and 2 are interchanged.

Row 1 is multiplied by k.

How does the row operation affect the determinant?

It multiplies the determinant by k.

It changes the sign of the determinant.

It increases the determinant by k.

It does not change the determinant.

3.1.22

Explore the effect of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

	a	b
	c	d

,

	a+kc	b+kd
	c	d

What is the elementary row operation?

Rows 1 and 2 are interchanged.

Row 1 is replaced with the sum of itself and k times row 2.

Row 2 is replaced with the sum of itself and k times row 1.

Row 1 is multiplied by k.

Row 2 is multiplied by k.

How does the row operation affect the determinant?

It increases the determinant by k.

It changes the sign of the determinant.

It multiplies the determinant by k.

It does not change the determinant.

3.1.23

Explore the effect of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

	3	6	5
	9	7	8
	a	b	c

,

	a	b	c
	9	7	8
	3	6	5

What is the elementary row operation?

Row

1

is replaced with the sum of rows

1

and

2.

Row

1

is replaced with the sum of rows

1

and

3.

Row

3

is replaced with the sum of rows

2

and

3.

Row

3

is replaced with the sum of rows

1

and

3.

Rows

1

and

3

are interchanged.

Rows

1

and

2

are interchanged.

Rows

2

and

3

are interchanged.

How does the row operation affect the determinant?

It changes the sign of the determinant.

It increases the determinant by 1.

It multiplies the determinant by 2.

It does not change the determinant.

3.1.23

Explore the effect of an elementary row operation on the determinant of a matrix. State the row operation and describe how it affects the determinant.

	4	7	1
	a	b	c
	2	6	5

,

	a	b	c
	4	7	1
	2	6	5

What is the elementary row operation?

Row

1

is replaced with the sum of rows

1

and

2.

Rows

2

and

3

are interchanged.

Row

2

is replaced with the sum of rows

1

and

2.

Row

1

is replaced with the sum of rows

1

and

3.

Rows

1

and

2

are interchanged.

Row

2

is replaced with the sum of rows

2

and

3.

Rows

1

and

3

are interchanged.

How does the row operation affect the determinant?

It changes the sign of the determinant.

It multiplies the determinant by 2.

It increases the determinant by 1.

It does not change the determinant.

3.1.26

Compute the determinant of the following elementary matrix.

	0	1	0
	1	0	0
	0	0	1

0	1	0
1	0	0
0	0	1

=−1

(Simplify your answer.)

3.1.37

Let

A=

	1	2
	7	9

.

Write

4A.

Is

det(4A)

equal to

4det(A)?

4A=

	4	8
	28	36

(Type an integer or decimal for each matrix element.)

Select the correct choice below and fill in the answer box(es) to complete your choice.

Yes,

det(4A)

is equal to

4det(A).

The value of both expressions is

.

No,

det(4A)

is not equal to

4det(A).

The value of

det(4A)

is

−80,

whereas the value of

4det(A)

is

−20.

3.1.37

Let

A=

	4	6
	5	2

.

Write

4A.

Is

det(4A)

equal to

4det(A)?

4A=

	16	24
	20	8

(Type an integer or decimal for each matrix element.)

Select the correct choice below and fill in the answer box(es) to complete your choice.

No,

det(4A)

is not equal to

4det(A).

The value of

det(4A)

is

−352,

whereas the value of

4det(A)

is

−88.

Yes,

det(4A)

is equal to

4det(A).

The value of both expressions is

.

3.1.37

Let

A=

	5	1
	8	9

.

Write

4A.

Is

det(4A)

equal to

4det(A)?

4A=

	20	4
	32	36

(Type an integer or decimal for each matrix element.)

Select the correct choice below and fill in the answer box(es) to complete your choice.

Yes,

det(4A)

is equal to

4det(A).

The value of both expressions is

.

No,

det(4A)

is not equal to

4det(A).

The value of

det(4A)

is

592,

whereas the value of

4det(A)

is

148.

3.1.39

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

An

n×n

determinant is defined by determinants of

(n−1)×(n−1)

submatrices.

Choose the correct answer below.

The statement is true. The determinant of an

n×n

matrix A can be computed by a cofactor expansion across any row or down any column. Each term in any such expansion includes a cofactor that involves the determinant of a submatrix of size

(n−1)×(n−1).

The statement is false. Although determinants of

(n−1)×(n−1)

submatrices can be used to find

n×n

determinants, they are not involved in the definition of

n×n

determinants.

The statement is false. An

n×n

determinant is defined by determinants of

(n−1)×(n−1)

submatrices only when

n>3.

Determinants of

1×1,

2×2,

and

3×3

matrices are defined separately.

The statement is true. The determinant of an

n×n

matrix A can be computed by a cofactor expansion along either diagonal. Each term in any such expansion includes a cofactor that involves the determinant of a submatrix of size

(n−1)×(n−1).

3.1.39

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

An

n×n

determinant is defined by determinants of

(n−1)×(n−1)

submatrices.

Choose the correct answer below.

The statement is false. An

n×n

determinant is defined by determinants of

(n−1)×(n−1)

submatrices only when

n>3.

Determinants of

1×1,

2×2,

and

3×3

matrices are defined separately.

The statement is true. The determinant of an

n×n

matrix A can be computed by a cofactor expansion along either diagonal. Each term in any such expansion includes a cofactor that involves the determinant of a submatrix of size

(n−1)×(n−1).

The statement is false. Although determinants of

(n−1)×(n−1)

submatrices can be used to find

n×n

determinants, they are not involved in the definition of

n×n

determinants.

The statement is true. The determinant of an

n×n

matrix A can be computed by a cofactor expansion across any row or down any column. Each term in any such expansion includes a cofactor that involves the determinant of a submatrix of size

(n−1)×(n−1).

3.1.40

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its ith row and jth column.

Choose the correct answer below.

The statement is true. It is the definition of the (i,j)-cofactor of a matrix A.

The statement is false. The (i,j)-cofactor of A is the number

Cij=detAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

The statement is false. The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its jth row and ith column.

The statement is false. The (i,j)-cofactor of A is the number

Cij=(−1)i+jdetAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

3.1.40

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its ith row and jth column.

Choose the correct answer below.

The statement is false. The (i,j)-cofactor of A is the number

Cij=(−1)i+jdetAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

The statement is false. The (i,j)-cofactor of A is the number

Cij=detAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

The statement is false. The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its jth row and ith column.

The statement is true. It is the definition of the (i,j)-cofactor of a matrix A.

3.1.40

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its ith row and jth column.

Choose the correct answer below.

The statement is false. The (i,j)-cofactor of a matrix A is the matrix

Aij

obtained by deleting from A its jth row and ith column.

The statement is false. The (i,j)-cofactor of A is the number

Cij=(−1)i+jdetAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

The statement is true. It is the definition of the (i,j)-cofactor of a matrix A.

The statement is false. The (i,j)-cofactor of A is the number

Cij=detAij,

where

Aij

is the submatrix obtained by deleting from A its ith row and jth column.

3.1.42

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The determinant of a triangular matrix is the sum of the entries on the main diagonal.

Choose the correct answer below.

The statement is true. Cofactor expansion along the row (or column) with the most zeros of a triangular matrix produces a determinant equal to the sum of the entries along the main diagonal.

The statement is false. The determinant of a triangular matrix is the product of the entries along the main diagonal.

The statement is false. The determinant of a matrix is the arithmetic mean of the entries along the main diagonal.

The statement is true. The determinant of A is the following finite series.

det A=∑j=1n(−1)1+ja1jdet A1j

In a triangular matrix, this series simplifies to the sum of the entries along the main diagonal.

3.1.42

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The determinant of a triangular matrix is the sum of the entries on the main diagonal.

Choose the correct answer below.

The statement is false. The determinant of a triangular matrix is the product of the entries along the main diagonal.

The statement is true. The determinant of A is the following finite series.

det A=∑j=1n(−1)1+ja1jdet A1j

In a triangular matrix, this series simplifies to the sum of the entries along the main diagonal.

The statement is true. Cofactor expansion along the row (or column) with the most zeros of a triangular matrix produces a determinant equal to the sum of the entries along the main diagonal.

The statement is false. The determinant of a matrix is the arithmetic mean of the entries along the main diagonal.

3.1.42

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The determinant of a triangular matrix is the sum of the entries on the main diagonal.

Choose the correct answer below.

The statement is false. The determinant of a matrix is the arithmetic mean of the entries along the main diagonal.

The statement is true. The determinant of A is the following finite series.

det A=∑j=1n(−1)1+ja1jdet A1j

In a triangular matrix, this series simplifies to the sum of the entries along the main diagonal.

The statement is false. The determinant of a triangular matrix is the product of the entries along the main diagonal.

The statement is true. Cofactor expansion along the row (or column) with the most zeros of a triangular matrix produces a determinant equal to the sum of the entries along the main diagonal.

3.2.1

State which property of determinants is illustrated in this equation.

−6	0	2
−18	−2	−6
−5	−7	9

=−

−18	−2	−6
−6	0	2
−5	−7	9

Choose the correct answer below.

If one row of A is multiplied by k to produce B, then

det B=k•det A.

If A and B are square matrices, then det AB=(det A)(det B).

If two rows of A are interchanged to produce B, then det B=−det A.

If a multiple of one row of A is added to another row to produce matrix B, then det B=det A.

3.2.1

State which property of determinants is illustrated in this equation.

2	−4	−9
−4	2	6
5	0	−1

=−

−4	2	6
2	−4	−9
5	0	−1

Choose the correct answer below.

If two rows of A are interchanged to produce B, then det B=−det A.

If A and B are square matrices, then det AB=(det A)(det B).

If one row of A is multiplied by k to produce B, then

det B=k•det A.

If a multiple of one row of A is added to another row to produce matrix B, then det B=det A.

3.2.2

The equation below illustrates a property of determinants. State the property.

3	−6	9
3	5	−5
1	3	3

=3

1	−2	3
3	5	−5
1	3	3

Multiplying a row by 3 divides the determinant by 3.

Multiplying a row by 3 multiplies the determinant by 3.

Factoring 3 out of the entire matrix divides the determinant by 3.

Dividing a row by 3 multiplies the determinant by 3.

3.2.2

The equation below illustrates a property of determinants. State the property.

3	−6	9
3	5	−5
1	3	3

=3

1	−2	3
3	5	−5
1	3	3

Multiplying a row by 3 multiplies the determinant by 3.

Multiplying a row by 3 divides the determinant by 3.

Dividing a row by 3 multiplies the determinant by 3.

Factoring 3 out of the entire matrix divides the determinant by 3.

3.2.4

State which property of determinants is illustrated in this equation.

4	4	3
12	−9	8
−4	5	−1

=

4	4	3
0	−21	−1
−4	5	−1

Choose the correct answer below.

If two rows of A are interchanged to produce B, then

det B=−det A.

If A and B are square matrices, then

det AB=(det A)(det B).

If a multiple of one row of A is added to another row to produce matrix B, then

det B=det A.

If one row of A is multiplied by k to produce B, then

det B=k•det A.

3.2.4

State which property of determinants is illustrated in this equation.

−8	−5	−2
−24	4	5
0	−2	−7

=

−8	−5	−2
0	19	11
0	−2	−7

Choose the correct answer below.

If two rows of A are interchanged to produce B, then

det B=−det A.

If a multiple of one row of A is added to another row to produce matrix B, then

det B=det A.

If one row of A is multiplied by k to produce B, then

det B=k•det A.

If A and B are square matrices, then

det AB=(det A)(det B).

3.2.5

Find the determinant by row reduction to echelon form.

1	5	−7
1	6	5
−1	−4	7

Use row operations to reduce the matrix to echelon form.

1	5	−7
1	6	5
−1	−4	7

~

	1	0	0
	0	1	0
	0	0	1

Find the determinant of the given matrix.

1	5	−7
1	6	5
−1	−4	7

=−12

(Simplify your answer.)

3.2.5

Find the determinant by row reduction to echelon form.

1	5	−7
−1	−4	−5
2	8	7

Use row operations to reduce the matrix to echelon form.

1	5	−7
−1	−4	−5
2	8	7

~

	1	0	0
	0	1	0
	0	0	1

Find the determinant of the given matrix.

1	5	−7
−1	−4	−5
2	8	7

=−3

(Simplify your answer.)

3.2.5

Find the determinant by row reduction to echelon form.

1	5	−7
1	6	5
−1	−4	7

Use row operations to reduce the matrix to echelon form.

1	5	−7
1	6	5
−1	−4	7

~

	1	0	0
	0	1	0
	0	0	1

Find the determinant of the given matrix.

1	5	−7
1	6	5
−1	−4	7

=−12

(Simplify your answer.)

3.2.7

Find the determinant by row reduction to echelon form.

1	−1	−3	0
4	−3	3	2
3	−1	−4	2
−1	3	8	2

Use row operations to reduce the matrix to echelon form.

	1	−1	−3	0
	4	−3	3	2
	3	−1	−4	2
	−1	3	8	2

~

	1	0	0	2625
	0	1	0	45
	0	0	1	225
	0	0	0	0

Find the determinant of the given matrix.

1	−1	−3	0
4	−3	3	2
3	−1	−4	2
−1	3	8	2

=0

(Simplify your answer.)

3.2.15

Find the determinant below, where

a	b	c
d	e	f
g	h	i

=6.

a	b	c
d	e	f
4g	4h	4i

a	b	c
d	e	f
4g	4h	4i

=24

(Simplify your answer.)

3.2.15

Find the determinant below, where

a	b	c
d	e	f
g	h	i

=8.

a	b	c
d	e	f
5g	5h	5i

a	b	c
d	e	f
5g	5h	5i

=40

(Simplify your answer.)

3.2.20

If  

a	b	c
d	e	f
g	h	i

=20,

find the determinant of

a	b	c
g	h	i
d	e	f

.

The determinant is

−20.

3.2.20

If  

a	b	c
d	e	f
g	h	i

=−9,

find the determinant of

a	b	c
g	h	i
d	e	f

.

The determinant is

9.

3.2.20

If  

a	b	c
d	e	f
g	h	i

=4,

find the determinant of

a	b	c
g	h	i
d	e	f

.

The determinant is

−4.

3.2.21

Use determinants to find out if the matrix is invertible.

5	0	−1
1	−3	−2
0	5	3

The determinant of the matrix is

0.

(Simplify your answer.)

Is the matrix invertible? Choose the correct answer below.

The matrix is not invertible.

The matrix is invertible.

3.2.22

Use determinants to find out if the matrix is invertible.

−20	−5	−1
−4	15	−2
0	−25	3

The determinant of the matrix is

−60.

(Simplify your answer.)

Is the matrix invertible?

The matrix is not invertible because the determinant is not zero.

The matrix is invertible because the determinant of the matrix is not zero.

The matrix is invertible because the determinant is defined.

The matrix is not invertible because the determinant is defined.

3.2.22

Use determinants to find out if the matrix is invertible.

−10	−3	2
−2	9	4
0	−15	−6

The determinant of the matrix is

36.

(Simplify your answer.)

Is the matrix invertible?

The matrix is invertible because the determinant is defined.

The matrix is not invertible because the determinant is not zero.

The matrix is not invertible because the determinant is defined.

The matrix is invertible because the determinant of the matrix is not zero.

3.2.22

Use determinants to find out if the matrix is invertible.

20	1	−3
4	−3	−6
0	5	9

The determinant of the matrix is

−36.

(Simplify your answer.)

Is the matrix invertible?

The matrix is not invertible because the determinant is defined.

The matrix is invertible because the determinant of the matrix is not zero.

The matrix is invertible because the determinant is defined.

The matrix is not invertible because the determinant is not zero.

3.2.25

Use determinants to decide if the set of vectors is linearly independent.

	−3
	−3
	1

,

	2
	4
	−3

,

	−3
	0
	−4

The determinant of the matrix whose columns are the given vectors is

9.

(Simplify your answer.)

Is the set of vectors linearly independent?

The set of vectors is linearly dependent, because the determinant exists.

The set of vectors is linearly dependent, because the determinant is not zero.

The set of vectors is linearly independent, because the determinant is not zero.

The set of vectors is linearly independent, because the determinant exists.

3.2.25

Use determinants to decide if the set of vectors is linearly independent.

	−2
	7
	3

,

	1
	−6
	−2

,

	−2
	0
	6

The determinant of the matrix whose columns are the given vectors is

22.

(Simplify your answer.)

Is the set of vectors linearly independent?

The set of vectors is linearly independent, because the determinant is not zero.

The set of vectors is linearly dependent, because the determinant exists.

The set of vectors is linearly dependent, because the determinant is not zero.

The set of vectors is linearly independent, because the determinant exists.

3.2.25

Use determinants to decide if the set of vectors is linearly independent.

	2
	2
	6

,

	−3
	−1
	2

,

	2
	0
	1

The determinant of the matrix whose columns are the given vectors is

24.

(Simplify your answer.)

Is the set of vectors linearly independent?

The set of vectors is linearly independent, because the determinant exists.

The set of vectors is linearly independent, because the determinant is not zero.

The set of vectors is linearly dependent, because the determinant exists.

The set of vectors is linearly dependent, because the determinant is not zero.

3.2.28

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If det A is zero, then two rows or two columns are the same, or a row or a column is zero.

Choose the correct answer below.

The statement is false. If

A=

	2	6
	1	3

,

then det

A=0

and the rows and columns are all distinct and not full of zeros.

The statement is true. If det A is zero, then the columns of A are linearly independent. If one column is zero, or two columns are the same, then the columns are linearly dependent.

The statement is true. If

A=

	2	3
	2	3

and

B=

	1	2
	0	0

,

then det

A=0

and det

B=0.

The statement is false. The determinant depends on the columns of A. It is possible for two rows to be the same and for the determinant to be nonzero.

3.2.28

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If det A is zero, then two rows or two columns are the same, or a row or a column is zero.

Choose the correct answer below.

The statement is false. If

A=

	2	6
	1	3

,

then det

A=0

and the rows and columns are all distinct and not full of zeros.

The statement is false. The determinant depends on the columns of A. It is possible for two rows to be the same and for the determinant to be nonzero.

The statement is true. If det A is zero, then the columns of A are linearly independent. If one column is zero, or two columns are the same, then the columns are linearly dependent.

The statement is true. If

A=

	2	3
	2	3

and

B=

	1	2
	0	0

,

then det

A=0

and det

B=0.

3.2.28

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If det A is zero, then two rows or two columns are the same, or a row or a column is zero.

Choose the correct answer below.

The statement is true. If

A=

	2	3
	2	3

and

B=

	1	2
	0	0

,

then det

A=0

and det

B=0.

The statement is false. If

A=

	2	6
	1	3

,

then det

A=0

and the rows and columns are all distinct and not full of zeros.

The statement is false. The determinant depends on the columns of A. It is possible for two rows to be the same and for the determinant to be nonzero.

The statement is true. If det A is zero, then the columns of A are linearly independent. If one column is zero, or two columns are the same, then the columns are linearly dependent.

3.2.29

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If the columns of A are linearly dependent, then det

A=0.

Choose the correct answer below.

The statement is false. If det

A=0,

then A is invertible.

The statement is false. The columns of I are linearly dependent, yet det I

=1.

The statement is true. If the columns of A are linearly dependent, then one of the columns is equal to another.

The statement is true. If the columns of A are linearly dependent, then A is not invertible.

3.2.29

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If the columns of A are linearly dependent, then det

A=0.

Choose the correct answer below.

The statement is false. If det

A=0,

then A is invertible.

The statement is true. If the columns of A are linearly dependent, then A is not invertible.

The statement is false. The columns of I are linearly dependent, yet det I

=1.

The statement is true. If the columns of A are linearly dependent, then one of the columns is equal to another.

3.2.29

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If the columns of A are linearly dependent, then det

A=0.

Choose the correct answer below.

The statement is true. If the columns of A are linearly dependent, then A is not invertible.

The statement is false. The columns of I are linearly dependent, yet det I

=1.

The statement is true. If the columns of A are linearly dependent, then one of the columns is equal to another.

The statement is false. If det

A=0,

then A is invertible.

3.2.32

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The determinant of A is the product of the pivots in any echelon form U of A, multiplied by

(−1)r,

where r is the number of row interchanges made during row reduction from A to U.

Choose the correct answer below.

The statement is false. Reduction to an echelon form may also include scaling a row by a nonzero constant, which can change the value of the determinant.

The statement is false. The determinant is the product of the number of pivots and

(−1)r.

The statement is true. If

A=

	2	3
	0	4

,

then det

A=8.

The statement is true. The determinant is the product of the entries on the diagonal and the pivots are all on the diagonal.

3.2.32

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The determinant of A is the product of the pivots in any echelon form U of A, multiplied by

(−1)r,

where r is the number of row interchanges made during row reduction from A to U.

Choose the correct answer below.

The statement is true. The determinant is the product of the entries on the diagonal and the pivots are all on the diagonal.

The statement is false. The determinant is the product of the number of pivots and

(−1)r.

The statement is false. Reduction to an echelon form may also include scaling a row by a nonzero constant, which can change the value of the determinant.

The statement is true. If

A=

	2	3
	0	4

,

then det

A=8.

3.2.33

Let A and B be

n×n

matrices. Determine whether the statement below is true or false. Justify the answer.

det(A+B)=det

A+det

B

Choose the correct answer below.

The statement is true. If

A=

	2	0
	1	0

and

B=

	3	0
	5	0

,

then

det(A+B)=0

and

det A+det B=0.

The statement is true. Determinants are linear transformations.

The statement is false. If

A=

	1	0
	0	1

and

B=

	−1	0
	0	−1

,

then

det(A+B)=0

and det

A+det

B=2.

The statement is false.

det(A+B)=(det

A)(det B)

3.2.35

Compute

det B4

where

B=

1	0	2
1	2	3
2	3	1

.

             

det B4=6561

(Simplify your answer.)

3.2.35

Compute

det B4

where

B=

1	0	2
2	1	2
1	2	1

.

             

det B4=81

(Simplify your answer.)

3.2.35

Compute

det B4

where

B=

2	0	1
2	1	3
1	2	1

.

             

det B4=2401

(Simplify your answer.)

3.2.38

Suppose that A is a square matrix such that det

A3=0.

Explain why A cannot be invertible.

Since the desired conclusion is that A is not invertible, it will be helpful to identify any conditions that are logically equivalent to the condition that A is not invertible. Which of the following conditions is logically equivalent to the condition that A is not invertible?

det

AB=(det

A)(det B)

det

A=0

det

A≠0

det

AT=det

A

What property can be applied to the left side of the equation det

A3=0

in order to yield the desired result?

det

A≠0

if and only if A is invertible.

det

A3=(det

A)(det A)(det A)

det

AT=det

A

If two rows of A are interchanged to produce B, then det

B=−det

A.

What should be done next to establish that A is not invertible? Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or a decimal.)

By the zero factor property, det

A=0.

Because det

B=,

det

A=.

By the zero factor property, det

AT=.

Because det

AB=,

det

A=.

Thus, A is not invertible.

3.2.38

Suppose that A is a square matrix such that det

A3=0.

Explain why A cannot be invertible.

Since the desired conclusion is that A is not invertible, it will be helpful to identify any conditions that are logically equivalent to the condition that A is not invertible. Which of the following conditions is logically equivalent to the condition that A is not invertible?

det

A≠0

det

AB=(det

A)(det B)

det

AT=det

A

det

A=0

What property can be applied to the left side of the equation det

A3=0

in order to yield the desired result?

If two rows of A are interchanged to produce B, then det

B=−det

A.

det

AT=det

A

det

A≠0

if and only if A is invertible.

det

A3=(det

A)(det A)(det A)

What should be done next to establish that A is not invertible? Select the correct choice below and fill in the answer box(es) within your choice.

(Type an integer or a decimal.)

By the zero factor property, det

A=0.

By the zero factor property, det

AT=.

Because det

B=,

det

A=.

Because det

AB=,

det

A=.

Thus, A is not invertible.

3.2.51

Suppose A is an

n×n

matrix and a computer suggests that

det A=6

and

det A−1=1.

Should you trust these answers? Why or why not?

Choose the correct answer below.

Yes, you should trust the answers because both determinants are non-zero and

det A−1

is the same sign as

det A.

No, you should not trust the answers because

det A×det A−1

should always equal 1.

No, you should not trust the answers because

det A×det A−1

should always equal

−1.

No, you should not trust the answers because

det A

should always be the opposite sign of

det A−1.

3.2.51

Suppose A is an

n×n

matrix and a computer suggests that

det A=1

and

det A−1=7.

Should you trust these answers? Why or why not?

Choose the correct answer below.

No, you should not trust the answers because

det A

should always be the opposite sign of

det A−1.

Yes, you should trust the answers because both determinants are non-zero and

det A−1

is the same sign as

det A.

No, you should not trust the answers because

det A×det A−1

should always equal

−1.

No, you should not trust the answers because

det A×det A−1

should always equal 1.

3.3.1

Use Cramer's rule to compute the solutions of the system.

 2x1+3x2=−2

 2x1+7x2=−26

What is the solution of the system?

x1=8

x2=−6

3.3.1

Use Cramer's rule to compute the solutions of the system.

 4x1+3x2=−5

 5x1+6x2=5

What is the solution of the system?

x1=−5

x2=5

3.3.1

Use Cramer's rule to compute the solutions of the system.

 2x1+3x2=3

 5x1+7x2=5

What is the solution of the system?

x1=−6

x2=5

3.3.3

Use Cramer's rule to compute the solutions of the system.

−2x1+7x2=	6
4x1+7x2=	5

What is the solution of the system?

x1=−16   

x2=1721

(Type integers or simplified fractions.)

3.3.7

Determine the values of the parameter s for which the system has a unique solution, and describe the solution.

4sx1+7x2	=	3
8x1+3sx2	=	−3

Choose the correct answer below.

s≠±143;

x1=3(−s−2)3s2−14;

x2=3(3s+7)43s2−14

s≠0;

x1=3(3s+7)43s2−14;

x2=3(−s−2)3s2−14

s≠0;

x1=3(−s−2)3s2−14;

x2=3(3s+7)43s2−14

s≠±143;

x1=3(3s+7)43s2−14;

x2=3(−s−2)3s2−14

3.3.7

Determine the values of the parameter s for which the system has a unique solution, and describe the solution.

3sx1+4x2	=	5
9x1+4sx2	=	−4

Choose the correct answer below.

s≠0;

x1=5s+43s2−3;

x2=−4s−154s2−3

s≠0;

x1=−4s−154s2−3;

x2=5s+43s2−3

s≠±3;

x1=−4s−154s2−3;

x2=5s+43s2−3

s≠±3;

x1=5s+43s2−3;

x2=−4s−154s2−3

3.3.7

Determine the values of the parameter s for which the system has a unique solution, and describe the solution.

2sx1+4x2	=	5
9x1+3sx2	=	−4

Choose the correct answer below.

s≠0;

x1=−8s−456s2−6;

x2=15s+166s2−6

s≠±6;

x1=15s+166s2−6;

x2=−8s−456s2−6

s≠±6;

x1=−8s−456s2−6;

x2=15s+166s2−6

s≠0;

x1=15s+166s2−6;

x2=−8s−456s2−6

3.3.11

Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix.

A=

0	−3	−1
3	0	0
−2	2	1

The adjugate of the given matrix is adj A=

	0	1	0
	−3	−2	−3
	6	6	9

.

(Type an integer or simplified fraction for each matrix element.)

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	0	13	0
	−1	−23	−1
	2	2	3

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

3.3.11

Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix.

A=

0	−2	−1
3	0	0
−2	1	1

The adjugate of the given matrix is adj A=

	0	1	0
	−3	−2	−3
	3	4	6

.

(Type an integer or simplified fraction for each matrix element.)

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	0	13	0
	−1	−23	−1
	1	43	2

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

3.3.11

Compute the adjugate of the given matrix, and then use the Inverse Formula to give the inverse of the matrix.

A=

0	−3	−1
6	0	0
−2	2	1

The adjugate of the given matrix is adj A=

	0	1	0
	−6	−2	−6
	12	6	18

.

(Type an integer or simplified fraction for each matrix element.)

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The inverse matrix is

A−1=

	0	16	0
	−1	−13	−1
	2	1	3

.

(Type an integer or simplified fraction for each matrix element.)

The matrix is not invertible.

3.3.18

Suppose that all the entries in A are integers and det

A=1.

Explain why all the entries in

A−1

are integers.

Choose the correct answer below.

Each cofactor in A is an integer because it is just a sum of products of entries of A. Hence all the entries in adj A are integers. Since det

A=1,

the inverse formula shows that all the entries in

A−1

are integers.

The inverse of A is always equal to the reciprocal of the determinant multiplied by matrix A. Since det

A=1,

the reciprocal is also equal to one, so the inverse of A is equal to matrix A.

Each cofactor in A is a fraction with the same denominator because it is just a sum of quotient of entries of A. All of the fractions sum to an integer. Since det

A=1,

the inverse formula shows that all the entries in

A−1

are integers.

The inverse of A is always equal to the negative reciprocal of the determinant multiplied by matrix A. Since det

A=1,

the reciprocal is also equal to one, so each component of the inverse matrix is equal to the negative of the components of matrix A.

3.3.18

Suppose that all the entries in A are integers and det

A=1.

Explain why all the entries in

A−1

are integers.

Choose the correct answer below.

The inverse of A is always equal to the reciprocal of the determinant multiplied by matrix A. Since det

A=1,

the reciprocal is also equal to one, so the inverse of A is equal to matrix A.

Each cofactor in A is an integer because it is just a sum of products of entries of A. Hence all the entries in adj A are integers. Since det

A=1,

the inverse formula shows that all the entries in

A−1

are integers.

Each cofactor in A is a fraction with the same denominator because it is just a sum of quotient of entries of A. All of the fractions sum to an integer. Since det

A=1,

the inverse formula shows that all the entries in

A−1

are integers.

The inverse of A is always equal to the negative reciprocal of the determinant multiplied by matrix A. Since det

A=1,

the reciprocal is also equal to one, so each component of the inverse matrix is equal to the negative of the components of matrix A.

5.1.2

Is

λ=−2

an eigenvalue of

A=

2	−2
−4	0

?

Why or why not?

Choose the correct answer below.

Yes,

λ

is an eigenvalue of A because

(A−λI)

is invertible.

Yes,

λ

is an eigenvalue of A because

Ax=λx

has a nontrivial solution.

Yes,

λ

is an eigenvalue of A because

λAx=0

only has the trivial solution.

No,

λ

is not an eigenvalue of A because

Ax=λx

has a nontrivial solution.

No,

λ

is not an eigenvalue of A because

λAx=0

only has the trivial solution.

No,

λ

is not an eigenvalue of A because

Ax=λx

only has the trivial solution.

5.1.2

Is

λ=−7

an eigenvalue of

A=

−9	5
−2	−2

?

Why or why not?

Choose the correct answer below.

No,

λ

is not an eigenvalue of A because

λAx=0

only has the trivial solution.

No,

λ

is not an eigenvalue of A because

Ax=λx

has a nontrivial solution.

Yes,

λ

is an eigenvalue of A because

λAx=0

only has the trivial solution.

Yes,

λ

is an eigenvalue of A because

(A−λI)

is invertible.

Yes,

λ

is an eigenvalue of A because

Ax=λx

has a nontrivial solution.

No,

λ

is not an eigenvalue of A because

Ax=λx

only has the trivial solution.

5.1.4

Is

v=

1

1

an eigenvector of

A=

−9	5
−2	−2

?

If so, find the eigenvalue.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes, v is an eigenvector of A. The eigenvalue is

λ=−4.

No, v is not an eigenvector of A.

5.1.4

Is

v=

−2

1

an eigenvector of

A=

−1	4
3	3

?

If so, find the eigenvalue.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes, v is an eigenvector of A. The eigenvalue is

λ=−3.

No, v is not an eigenvector of A.

5.1.6

Is

v=

0

1

0

an eigenvector of

A=

4	0	1
−2	1	0
−2	0	1

?

If so, find the eigenvalue.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes, v is an eigenvector of A. The eigenvalue is

λ=1.

No, v is not an eigenvector of A.

5.1.6

Is

v=

−1

5

2

an eigenvector of

A=

−1	0	−2
2	5	−4
0	2	−2

?

If so, find the eigenvalue.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes, v is an eigenvector of A. The eigenvalue is

λ=3.

No, v is not an eigenvector of A.

5.1.6

Is

v=

3

−2

1

an eigenvector of

A=

−4	3	3
2	−3	−2
−1	0	−2

?

If so, find the eigenvalue.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes, v is an eigenvector of A. The eigenvalue is

λ=−5.

No, v is not an eigenvector of A.

5.1.7

Is

λ=8

an eigenvalue of

	7	0	−3
	2	7	5
	−3	4	3

?

If so, find one corresponding eigenvector.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes,

λ=8

is an eigenvalue of

	7	0	−3
	2	7	5
	−3	4	3

.

One corresponding eigenvector is

	−3
	−1
	1

.

(Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element.)

No,

λ=8

is not an eigenvalue of

	7	0	−3
	2	7	5
	−3	4	3

.

5.1.7

Is

λ=7

an eigenvalue of

	6	0	−3
	3	6	6
	−4	2	1

?

If so, find one corresponding eigenvector.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes,

λ=7

is an eigenvalue of

	6	0	−3
	3	6	6
	−4	2	1

.

One corresponding eigenvector is

	−3
	−3
	1

.

(Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element.)

No,

λ=7

is not an eigenvalue of

	6	0	−3
	3	6	6
	−4	2	1

.

5.1.7

Is

λ=5

an eigenvalue of

	4	0	−1
	2	4	3
	−4	3	−2

?

If so, find one corresponding eigenvector.

Select the correct choice below and, if necessary, fill in the answer box within your choice.

Yes,

λ=5

is an eigenvalue of

	4	0	−1
	2	4	3
	−4	3	−2

.

One corresponding eigenvector is

	−1
	1
	1

.

(Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element.)

No,

λ=5

is not an eigenvalue of

	4	0	−1
	2	4	3
	−4	3	−2

.

5.1.9

Find a basis for the eigenspace corresponding to each listed eigenvalue of A below.

A=

	3	0
	−1	4

,

λ=4,3

A basis for the eigenspace corresponding to

λ=4

is

	0
	1

.

(Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to

λ=3

is

	1
	1

.

(Use a comma to separate answers as needed.)

5.1.9

Find a basis for the eigenspace corresponding to each listed eigenvalue of A below.

A=

	1	0
	−1	2

,

λ=2,1

A basis for the eigenspace corresponding to

λ=2

is

	0
	1

.

(Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to

λ=1

is

	1
	1

.

(Use a comma to separate answers as needed.)

5.1.9

Find a basis for the eigenspace corresponding to each listed eigenvalue of A below.

A=

	1	0
	−4	5

,

λ=5,1

A basis for the eigenspace corresponding to

λ=5

is

	0
	1

.

(Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to

λ=1

is

	1
	1

.

(Use a comma to separate answers as needed.)

5.1.12

Find a basis for the eigenspace corresponding to each listed eigenvalue.

A=

	5	6
	−2	−2

,

λ=1,

2

A basis for the eigenspace corresponding to

λ=1

is

	−32
	1

.

(Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to

λ=2

is

	−2
	1

.

(Type a vector or list of vectors. Type an integer or simplified fraction for each matrix element. Use a comma to separate answers as needed.)

5.1.14

Find a basis for the eigenspace corresponding to the eigenvalue of A given below.

A=

	5	0	1
	−3	0	1
	−1	−1	4

,

λ=4

A basis for the eigenspace corresponding to

λ=4

is

	−1
	1
	1

.

(Use a comma to separate answers as needed.)

5.1.14

Find a basis for the eigenspace corresponding to the eigenvalue of A given below.

A=

	7	0	−2
	2	0	−10
	−3	3	15

,

λ=6

A basis for the eigenspace corresponding to

λ=6

is

	2
	−1
	1

.

(Use a comma to separate answers as needed.)

5.1.17

Find the eigenvalues of the matrix.

	0	0	0
	0	5	−8
	0	0	4

The eigenvalue(s) of the matrix is/are

0,4,5.

(Use a comma to separate answers as needed.)

5.1.19

For

A=

1	2	4
1	2	4
1	2	4

,

find one eigenvalue, with no calculation. Justify your answer.

Choose the correct answer below.

One eigenvalue of A is

λ=1.

This is because each row of A is equal to the product of 1 and the row above it.

One eigenvalue of A is

λ=2.

This is because each column of A is equal to the

product

of

2

and the column to the left of it.

One eigenvalue of A is

λ=1.

This is because

1

is one of the entries on the main diagonal of A, which are the eigenvalues of A.

One eigenvalue of A is

λ=0.

This is because the columns of A are linearly dependent, so the matrix is not invertible.

5.1.19

For

A=

3	7	11
3	7	11
3	7	11

,

find one eigenvalue, with no calculation. Justify your answer.

Choose the correct answer below.

One eigenvalue of A is

λ=0.

This is because the columns of A are linearly dependent, so the matrix is not invertible.

One eigenvalue of A is

λ=4.

This is because each column of A is equal to the

sum

of

4

and the column to the left of it.

One eigenvalue of A is

λ=1.

This is because each row of A is equal to the product of 1 and the row above it.

One eigenvalue of A is

λ=3.

This is because

3

is one of the entries on the main diagonal of A, which are the eigenvalues of A.

5.1.21

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A.

Choose the correct answer below.

The statement is true. If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A because the only solution to this equation is the trivial solution.

The statement is false. The equation

Ax=λx

is not used to determine eigenvalues. If

λAx=0

for some vector

x,

then

λ

is an eigenvalue of A.

The statement is false. The condition that

Ax=λx

for some vector

x

is not sufficient to determine if

λ

is an eigenvalue. The equation

Ax=λx

must have a nontrivial solution.

The statement is true. If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A by the definition of an eigenvalue.

5.1.21

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A.

Choose the correct answer below.

The statement is false. The condition that

Ax=λx

for some vector

x

is not sufficient to determine if

λ

is an eigenvalue. The equation

Ax=λx

must have a nontrivial solution.

The statement is true. If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A by the definition of an eigenvalue.

The statement is true. If

Ax=λx

for some vector

x,

then

λ

is an eigenvalue of A because the only solution to this equation is the trivial solution.

The statement is false. The equation

Ax=λx

is not used to determine eigenvalues. If

λAx=0

for some vector

x,

then

λ

is an eigenvalue of A.

5.1.22

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

Ax=λx

for some scalar

λ,

then

x

is an eigenvector of A.

Choose the correct answer below.

The statement is false. The equation

Ax=λx

is not used to determine eigenvectors. If

λAx=0

for some scalar

λ,

then

x

is an eigenvector of A.

The statement is false. The condition that

Ax=λx

for some scalar

λ

is not sufficient to determine if

x

is an eigenvector of A. The vector x must be nonzero.

The statement is true. If

Ax=λx

for some scalar

λ,

then

x

is an eigenvector of A because

λ

is an inverse of A.

The statement is true. If

Ax=λx

for some scalar

λ,

then

x

is an eigenvector of A because the only solution to this equation is the trivial solution.

5.1.23

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

A matrix A is not invertible if and only if 0 is an eigenvalue of A.

Choose the correct answer below.

The statement is false. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation

Ax=0x.

The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has nontrivial solutions if and only if A is invertible.

The statement is false. If 0 is an eigenvalue of A, then the equation

Ax=0x

has only the trivial solution. The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has only the trivial solution if and only if A is invertible.

The statement is true. If 0 is an eigenvalue of A, then the equation

Ax=0x

has only the trivial solution. The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has only the trivial solution if and only if A is not invertible.

The statement is true. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation

Ax=0x.

The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has nontrivial solutions if and only if A is not invertible.

5.1.23

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

A matrix A is not invertible if and only if 0 is an eigenvalue of A.

Choose the correct answer below.

The statement is true. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation

Ax=0x.

The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has nontrivial solutions if and only if A is not invertible.

The statement is false. If 0 is an eigenvalue of A, then the equation

Ax=0x

has only the trivial solution. The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has only the trivial solution if and only if A is invertible.

The statement is false. If 0 is an eigenvalue of A, then there are nontrivial solutions to the equation

Ax=0x.

The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has nontrivial solutions if and only if A is invertible.

The statement is true. If 0 is an eigenvalue of A, then the equation

Ax=0x

has only the trivial solution. The equation

Ax=0x

is equivalent to the equation

Ax=0,

and

Ax=0

has only the trivial solution if and only if A is not invertible.

5.1.24

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

A number c is an eigenvalue of A if and only if the equation

(A−cI)x=0

has a nontrivial solution.

Choose the correct answer below.

The statement is false. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has only the trivial solution, and

Ax=cx

and

(A−cI)x=0

are equivalent equations.

The statement is false. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has nontrivial solutions, and the equation

(A−cI)x=0

has no bearing on whether c is an eigenvalue.

The statement is true. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has only the trivial solution, and

Ax=cx

and

(A−cI)x=0

are inverse equations.

The statement is true. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has nontrivial solutions, and

Ax=cx

and

(A−cI)x=0

are equivalent equations.

5.1.24

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

A number c is an eigenvalue of A if and only if the equation

(A−cI)x=0

has a nontrivial solution.

Choose the correct answer below.

The statement is true. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has only the trivial solution, and

Ax=cx

and

(A−cI)x=0

are inverse equations.

The statement is true. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has nontrivial solutions, and

Ax=cx

and

(A−cI)x=0

are equivalent equations.

The statement is false. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has nontrivial solutions, and the equation

(A−cI)x=0

has no bearing on whether c is an eigenvalue.

The statement is false. A number c is an eigenvalue of A if and only if the equation

Ax=cx

has only the trivial solution, and

Ax=cx

and

(A−cI)x=0

are equivalent equations.

5.1.26

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

To find the eigenvalues of A, reduce A to echelon form.

Choose the correct answer below.

The statement is true. An echelon form of a matrix A displays the eigenvalues as pivots of A.

The statement is false. An echelon form of

A−1

displays the eigenvalues of a matrix A.

The statement is true. An echelon form of a matrix A displays the eigenvalues on the main diagonal of A.

The statement is false. An echelon form of a matrix A usually does not display the eigenvalues of A.

5.1.26

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

To find the eigenvalues of A, reduce A to echelon form.

Choose the correct answer below.

The statement is false. An echelon form of

A−1

displays the eigenvalues of a matrix A.

The statement is false. An echelon form of a matrix A usually does not display the eigenvalues of A.

The statement is true. An echelon form of a matrix A displays the eigenvalues as pivots of A.

The statement is true. An echelon form of a matrix A displays the eigenvalues on the main diagonal of A.

5.1.27

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

v1

and

v2

are linearly independent eigenvectors, then they correspond to distinct eigenvalues.

Choose the correct answer below.

The statement is true. If

v1

and

v2

are linearly independent eigenvectors, then they correspond to distinct eigenvalues, because eigenvectors must be nonzero.

The statement is true. If

v1

and

v2

are linearly independent eigenvectors, then they correspond to distinct eigenvalues, because all eigenvectors of a single eigenvalue are linearly dependent.

The statement is false.

There

may be linearly independent eigenvectors that both correspond to the same eigenvalue.

The statement is false. Every eigenvalue has an infinite number of corresponding eigenvectors.

5.1.28

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The eigenvalues of a matrix are on its main diagonal.

Choose the correct answer below.

The statement is true. The eigenvalues of a matrix are on its main diagonal because the main diagonal remains the same when the matrix is transposed, and a matrix and its transpose have the same eigenvalues.

The statement is false. The matrix must first be reduced to echelon form. The eigenvalues are on the main diagonal of the echelon form of the matrix.

The statement is true. The eigenvalues of a matrix are on its main diagonal because the main diagonal determines the pivots of the matrix, which are used to calculate the eigenvalues.

The statement is false. If the matrix is a triangular matrix, the values on the main diagonal are eigenvalues. Otherwise, the main diagonal may or may not contain eigenvalues.

5.1.30

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

An eigenspace of A is a null space of a certain matrix.

Choose the correct answer below.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(A−λI).

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all solutions x to the equation

Ax=λb,

which does not include the zero vector unless

b=0.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(λA−I).

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all the eigenvectors corresponding to an eigenvalue

λ,

and eigenvectors are by definition nonzero vectors, so the eigenspace does not include the zero

vector.

5.1.30

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

An eigenspace of A is a null space of a certain matrix.

Choose the correct answer below.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(λA−I).

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all the eigenvectors corresponding to an eigenvalue

λ,

and eigenvectors are by definition nonzero vectors, so the eigenspace does not include the zero

vector.

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all solutions x to the equation

Ax=λb,

which does not include the zero vector unless

b=0.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(A−λI).

5.1.30

A is an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

An eigenspace of A is a null space of a certain matrix.

Choose the correct answer below.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(A−λI).

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all the eigenvectors corresponding to an eigenvalue

λ,

and eigenvectors are by definition nonzero vectors, so the eigenspace does not include the zero

vector.

The statement is false. An eigenspace of A is not a null space of a certain matrix because an eigenspace consists of all solutions x to the equation

Ax=λb,

which does not include the zero vector unless

b=0.

The statement is true. An eigenspace of A corresponding to the eigenvalue

λ

is the null space of the matrix

(λA−I).

5.2.1

Find the characteristic polynomial and the eigenvalues of the matrix.

7	5
5	7

The characteristic polynomial is

λ2−14λ+24.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

2,12.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.1

Find the characteristic polynomial and the eigenvalues of the matrix.

10	2
2	10

The characteristic polynomial is

λ2−20λ+96.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

8,12.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.1

Find the characteristic polynomial and the eigenvalues of the matrix.

7	5
5	7

The characteristic polynomial is

λ2−14λ+24.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

2,12.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.6

Find the characteristic polynomial and the eigenvalues of the matrix.

	7	−6
	6	2

The characteristic polynomial is

λ2−9λ+50.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.7

Find the characteristic polynomial and the eigenvalues of the matrix.

	−5	−2
	7	−7

The characteristic polynomial is

λ2+12λ+49.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.7

Find the characteristic polynomial and the eigenvalues of the matrix.

	9	2
	−5	5

The characteristic polynomial is

λ2−14λ+55.

(Type an expression using

λ

as the variable. Type an exact answer, using radicals as needed.)

Select the correct choice below and, if necessary, fill in the answer box within your choice.

The real eigenvalue(s) of the matrix is/are

.

(Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. Type each answer only once.)

The matrix has no real eigenvalues.

5.2.9

Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for

3×3

determinants. [Note: Finding the characteristic polynomial of a

3×3

matrix is not easy to do with just row operations, because the variable

λ

is involved.]

	1	0	1
	−5	2	−4
	0	5	0

The characteristic polynomial is

−λ3+3λ2−22λ−5.

(Type an expression using

λ

as the variable.)

5.2.9

Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for

3×3

determinants. [Note: Finding the characteristic polynomial of a

3×3

matrix is not easy to do with just row operations, because the variable

λ

is involved.]

	1	0	−1
	5	2	−4
	0	7	0

The characteristic polynomial is

−λ3+3λ2−30λ−7.

(Type an expression using

λ

as the variable.)

5.2.9

Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for

3×3

determinants. [Note: Finding the characteristic polynomial of a

3×3

matrix is not easy to do with just row operations, because the variable

λ

is involved.]

	1	0	−1
	5	2	−4
	0	7	0

The characteristic polynomial is

−λ3+3λ2−30λ−7.

(Type an expression using

λ

as the variable.)

5.2.18

It can be shown that the algebraic multiplicity of an eigenvalue

λ

is always greater than or equal to the dimension of the eigenspace corresponding to

λ.

Find h in the matrix A below such that the eigenspace for

λ=5

is two-dimensional.

A=

	5	−2	8	−3
	0	3	h	0
	0	0	5	7
	0	0	0	1

The value of h for which the eigenspace for

λ=5

is two-dimensional is

h=8.

5.2.18

It can be shown that the algebraic multiplicity of an eigenvalue

λ

is always greater than or equal to the dimension of the eigenspace corresponding to

λ.

Find h in the matrix A below such that the eigenspace for

λ=7

is two-dimensional.

A=

	7	−4	8	5
	0	3	h	0
	0	0	7	4
	0	0	0	4

The value of h for which the eigenspace for

λ=7

is two-dimensional is

h=8.

5.2.18

It can be shown that the algebraic multiplicity of an eigenvalue

λ

is always greater than or equal to the dimension of the eigenspace corresponding to

λ.

Find h in the matrix A below such that the eigenspace for

λ=8

is two-dimensional.

A=

	8	−3	6	5
	0	5	h	0
	0	0	8	3
	0	0	0	1

The value of h for which the eigenspace for

λ=8

is two-dimensional is

h=6.

5.2.27

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

λ+5

is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A.

Choose the correct answer below.

The statement is true. If

λ+5

is a factor of the characteristic polynomial of A, then

λ+5

must be an entry on the main diagonal of

A−λI.

The statement is false. If

λ+5

is a factor of the characteristic polynomial of A, then

−5

is an eigenvalue of A. In order for 5 to be an eigenvalue of A, the characteristic polynomial would need to have a factor of

λ−5.

The statement is false. The factors of the characteristic polynomial are useful for determining the eigenvectors of a matrix, not its eigenvalues.

The statement is true. If

λ+5

is a factor of the characteristic polynomial of A, then

λ=5

is a root of the characteristic equation.

5.2.27

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

λ+5

is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A.

Choose the correct answer below.

The statement is true. If

λ+5

is a factor of the characteristic polynomial of A, then

λ+5

must be an entry on the main diagonal of

A−λI.

The statement is true. If

λ+5

is a factor of the characteristic polynomial of A, then

λ=5

is a root of the characteristic equation.

The statement is false. The factors of the characteristic polynomial are useful for determining the eigenvectors of a matrix, not its eigenvalues.

The statement is false. If

λ+5

is a factor of the characteristic polynomial of A, then

−5

is an eigenvalue of A. In order for 5 to be an eigenvalue of A, the characteristic polynomial would need to have a factor of

λ−5.

5.2.28

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

The multiplicity of a root r of the characteristic equation of A is called the algebraic multiplicity of r as an eigenvalue of A.

Choose the correct answer below.

The statement is true. It is the definition of the algebraic multiplicity of an eigenvalue of A.

The statement is false. The multiplicity of a root r of the characteristic equation of A is the number of eigenvectors corresponding to that root.

The statement is true. It is the definition of the multiplicity of a root of the characteristic equation of A.

The statement is false. The multiplicity of a root r of the characteristic equation of A is called the geometric multiplicity of r as an eigenvalue of A.

5.3.1

Let

A=PDP−1

and P and D as shown below. Compute

A4.

P=

1	2
2	3

,

D=

1	0
0	2

A4=

	61	−30
	90	−44

(Simplify your answer.)

5.3.1

Let

A=PDP−1

and P and D as shown below. Compute

A4.

P=

1	4
2	7

,

D=

1	0
0	3

A4=

	641	−320
	1120	−559

(Simplify your answer.)

5.3.1

Let

A=PDP−1

and P and D as shown below. Compute

A4.

P=

1	4
2	7

,

D=

1	0
0	3

A4=

	641	−320
	1120	−559

(Simplify your answer.)

5.3.5

Matrix A is factored in the form

PDP−1.

Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace.

A=

3	1	1
1	3	1
1	1	3

=

	1	1	2
	1	0	−2
	1	−1	0

5	0	0
0	2	0
0	0	2

	13	13	13
	13	13	−23
	16	−13	16

Select the correct choice below and fill in the answer boxes to complete your choice.

(Use a comma to separate vectors as needed.)

There is one distinct eigenvalue,

λ=.

A basis for the corresponding eigenspace is

.

In ascending order, the two distinct eigenvalues are

λ1=2

and

λ2=5.

Bases for the corresponding eigenspaces are

	1
	0
	−1

,

	2
	−2
	0

and

	1
	1
	1

,

respectively.

In ascending order, the three distinct eigenvalues are

λ1=,

λ2=,

and

λ3=.

Bases for the corresponding eigenspaces are

,

,

and

,

respectively.

5.3.5

Matrix A is factored in the form

PDP−1.

Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace.

A=

2	1	1
2	3	2
1	1	2

=

	1	2	2
	2	0	−2
	1	−2	0

5	0	0
0	1	0
0	0	1

	14	14	14
	18	18	−38
	14	−14	14

Select the correct choice below and fill in the answer boxes to complete your choice.

(Use a comma to separate vectors as needed.)

There is one distinct eigenvalue,

λ=.

A basis for the corresponding eigenspace is

.

In ascending order, the two distinct eigenvalues are

λ1=1

and

λ2=5.

Bases for the corresponding eigenspaces are

	2
	0
	−2

,

	2
	−2
	0

and

	1
	2
	1

,

respectively.

In ascending order, the three distinct eigenvalues are

λ1=,

λ2=,

and

λ3=.

Bases for the corresponding eigenspaces are

,

,

and

,

respectively.

5.3.5

Matrix A is factored in the form

PDP−1.

Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace.

A=

2	2	1
1	3	1
1	2	2

=

	2	1	2
	2	0	−1
	2	−1	0

5	0	0
0	1	0
0	0	1

	18	14	18
	14	12	−34
	14	−12	14

Select the correct choice below and fill in the answer boxes to complete your choice.

(Use a comma to separate vectors as needed.)

There is one distinct eigenvalue,

λ=.

A basis for the corresponding eigenspace is

.

In ascending order, the two distinct eigenvalues are

λ1=1

and

λ2=5.

Bases for the corresponding eigenspaces are

	1
	0
	−1

,

	2
	−1
	0

and

	2
	2
	2

,

respectively.

In ascending order, the three distinct eigenvalues are

λ1=,

λ2=,

and

λ3=.

Bases for the corresponding eigenspaces are

,

,

and

,

respectively.

5.3.7

Diagonalize the following matrix, if possible.

	1	0
	8	−1

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=,

D=

	1	0
	0	4

For

P=

	1	0
	4	1

,

D=

	1	0
	0	−1

For

P=,

D=

	8	0
	0	−8

The matrix cannot be diagonalized.

5.3.7

Diagonalize the following matrix, if possible.

	7	0
	8	−7

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	7	0
	4	1

,

D=

	7	0
	0	−7

For

P=,

D=

	7	0
	0	4

For

P=,

D=

	8	0
	0	−8

The matrix cannot be diagonalized.

5.3.7

Diagonalize the following matrix, if possible.

	8	0
	10	−8

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	8	0
	5	1

,

D=

	8	0
	0	−8

For

P=,

D=

	10	0
	0	−10

For

P=,

D=

	8	0
	0	5

The matrix cannot be diagonalized.

5.3.8

Diagonalize the following matrix.

	5	1
	0	5

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=,

D=

	1	0
	0	−1

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	5	0
	0	5

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	5	0
	0	1

.

(Type an integer or simplified fraction for each matrix element.)

The matrix cannot be diagonalized.

5.3.8

Diagonalize the following matrix.

	18	1
	0	18

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=,

D=

	1	0
	0	−1

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	18	0
	0	1

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	18	0
	0	18

.

(Type an integer or simplified fraction for each matrix element.)

The matrix cannot be diagonalized.

5.3.11

Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.

	10	17	−26
	6	23	−28
	6	18	−23

;

λ=1,4,5

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	1	3	2
	1	2	4
	1	2	3

,

D=

	1	0	0
	0	4	0
	0	0	5

.

(Simplify your answer.)

The matrix cannot be diagonalized.

5.3.11

Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.

	−5	14	−8
	−9	16	−6
	−9	11	−1

;

λ=1,4,5

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	1	2	1
	1	3	3
	1	3	4

,

D=

	1	0	0
	0	4	0
	0	0	5

.

(Simplify your answer.)

The matrix cannot be diagonalized.

5.3.11

Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix.

	8	12	−18
	4	18	−20
	4	13	−15

;

λ=2,4,5

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	1	3	2
	1	2	4
	1	2	3

,

D=

	2	0	0
	0	4	0
	0	0	5

.

(Simplify your answer.)

The matrix cannot be diagonalized.

5.3.18

Diagonalize the following matrix. One eigenvalue is

λ=2

and one eigenvector is (−3,3,3).

	4	6	−1
	−2	−4	1
	−2	−6	3

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=,

D=

	1	0	0
	0	1	0
	0	0	2

.

(Type an integer or simplified fraction for each matrix element.)

For

P=

	−1	−6	1
	1	2	0
	1	0	2

,

D=

	−1	0	0
	0	2	0
	0	0	2

.

(Type an integer or simplified fraction for each matrix element.)

The matrix cannot be diagonalized.

5.3.18

Diagonalize the following matrix. One eigenvalue is

λ=6

and one eigenvector is (−1,1,2).

	17	44	−11
	−11	−38	11
	−22	−88	28

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	−1	−4	1
	1	1	0
	2	0	1

,

D=

	−5	0	0
	0	6	0
	0	0	6

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	22	0	0
	0	22	0
	0	0	6

.

(Type an integer or simplified fraction for each matrix element.)

The matrix cannot be diagonalized.

5.3.18

Diagonalize the following matrix. One eigenvalue is

λ=4

and one eigenvector is (−3,2,2).

	−8	−12	3
	8	12	−2
	8	8	2

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

For

P=

	−3	1	−1
	2	0	1
	2	4	0

,

D=

	−2	0	0
	0	4	0
	0	0	4

.

(Type an integer or simplified fraction for each matrix element.)

For

P=,

D=

	−2	0	0
	0	−2	0
	0	0	4

.

(Type an integer or simplified fraction for each matrix element.)

The matrix cannot be diagonalized.

5.3.21

Let A, P, and D be

n×n

matrices. Determine whether the statement below is true or false. Justify the answer.

A is diagonalizable if

A=PDP−1

for some matrix D and some invertible matrix P.

Choose the correct answer below.

The statement is false. A must have n distinct eigenvalues for the matrix D and P to exist.

The statement is false. The symbol D does not automatically denote a diagonal matrix.

The statement is true. If

A=PDP−1,

then A is diagonalizable by definition.

The statement is true. The columns of P are the n linearly independent eigenvectors of A.

5.3.21

Let A, P, and D be

n×n

matrices. Determine whether the statement below is true or false. Justify the answer.

A is diagonalizable if

A=PDP−1

for some matrix D and some invertible matrix P.

Choose the correct answer below.

The statement is false. A must have n distinct eigenvalues for the matrix D and P to exist.

The statement is true. If

A=PDP−1,

then A is diagonalizable by definition.

The statement is false. The symbol D does not automatically denote a diagonal matrix.

The statement is true. The columns of P are the n linearly independent eigenvectors of A.

5.3.22

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

ℝn

has a basis of eigenvectors of A, then A is diagonalizable.

Choose the correct answer below.

The statement is true. All square matrices are diagonalizable.

The statement is false. The eigenvectors of A must also be linearly independent.

The statement is false. A must also have n distinct eigenvalues.

The statement is true. A is diagonalizable if and only if there are enough eigenvectors to form a basis of

ℝn.

5.3.22

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If

ℝn

has a basis of eigenvectors of A, then A is diagonalizable.

Choose the correct answer below.

The statement is false. A must also have n distinct eigenvalues.

The statement is true. All square matrices are diagonalizable.

The statement is true. A is diagonalizable if and only if there are enough eigenvectors to form a basis of

ℝn.

The statement is false. The eigenvectors of A must also be linearly independent.

5.3.23

Let A, P, and D be

n×n

matrices. Determine whether the statement below is true or false. Justify the answer.

A is diagonalizable if and only if A has n eigenvalues, counting multiplicities.

Choose the correct answer below.

The statement is true.

A=PDP−1,

where the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.

The statement is true. There are n corresponding eigenvectors that form a basis of

ℝn.

The statement is false. The eigenvalues of A may not produce enough eigenvectors to form a basis of

ℝn.

The statement is false. It also required that A is invertible.

5.3.23

Let A, P, and D be

n×n

matrices. Determine whether the statement below is true or false. Justify the answer.

A is diagonalizable if and only if A has n eigenvalues, counting multiplicities.

Choose the correct answer below.

The statement is true.

A=PDP−1,

where the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.

The statement is false. It also required that A is invertible.

The statement is false. The eigenvalues of A may not produce enough eigenvectors to form a basis of

ℝn.

The statement is true. There are n corresponding eigenvectors that form a basis of

ℝn.

5.3.24

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If A is diagonalizable, then A is invertible.

Choose the correct answer below.

The statement is false. Invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.

The statement is true. If A is invertible, then it has n distinct eigenvectors that form a basis of

ℝn.

The statement is true. If A is diagonalizable, then det(A) does not equal 0. Thus, A is invertible.

The statement is false. If A is diagonalizable, then

det(A−λI)=0

has a

solution.

Thus, A is not invertible.

5.3.24

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If A is diagonalizable, then A is invertible.

Choose the correct answer below.

The statement is false. If A is diagonalizable, then

det(A−λI)=0

has a

solution.

Thus, A is not invertible.

The statement is true. If A is diagonalizable, then det(A) does not equal 0. Thus, A is invertible.

The statement is true. If A is invertible, then it has n distinct eigenvectors that form a basis of

ℝn.

The statement is false. Invertibility depends on 0 not being an eigenvalue. A diagonalizable matrix may or may not have 0 as an eigenvalue.

5.3.25

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

A matrix A is diagonalizable if A has n eigenvectors.

Choose the correct answer below.

The statement is false. A diagonalizable matrix must have n linearly independent eigenvectors.

The statement is true. A diagonalizable matrix must have a minimum of n linearly independent eigenvectors.

The statement is true. A diagonalizable matrix must have more than one linearly independent eigenvector.

The statement is false. A matrix is diagonalizable if and only if it has

n−1

linearly independent eigenvectors.

5.3.26

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If A is diagonalizable, then A has n distinct eigenvalues.

Choose the correct answer below.

The statement is true. A diagonalizable matrix must have exactly n eigenvalues.

The statement is false. A diagonalizable matrix must have more than n eigenvalues.

The statement is false. A diagonalizable matrix can have fewer than n eigenvalues and still have n linearly independent eigenvectors.

The statement is true. A diagonalizable matrix must have n distinct eigenvalues.

5.3.28

Let A be an

n×n

matrix. Determine whether the statement below is true or false. Justify the answer.

If A is invertible, then A is diagonalizable.

Choose the correct answer below.

The statement is false. An invertible matrix may have fewer than n linearly independent eigenvectors, making it not diagonalizable.

The statement is true. If a matrix is invertible, then it has n linearly independent eigenvectors, making it diagonalizable.

The statement is false.

Invertible

matrices always have a maximum of n linearly independent eigenvectors, making it not diagonalizable.

The statement is true. A diagonalizable matrix is invertible, so an invertible matrix is diagonalizable.

5.3.30

A is a

3×3

matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

No. A matrix with 3 columns must have

unique eigenvalues in order to be diagonalizable.

No. The sum of the dimensions of the eigenspaces equals

2

and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.

Yes. One of the eigenspaces would have

unique eigenvectors. Since the eigenvector for the third eigenvalue would also be unique, A must be diagonalizable.

Yes. As long as the collection of eigenvectors spans

ℝ3,

A is diagonalizable.

5.3.30

A is a

3×3

matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

Yes. One of the eigenspaces would have

unique eigenvectors. Since the eigenvector for the third eigenvalue would also be unique, A must be diagonalizable.

No. A matrix with 3 columns must have

unique eigenvalues in order to be diagonalizable.

No. The sum of the dimensions of the eigenspaces equals

2

and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.

Yes. As long as the collection of eigenvectors spans

ℝ3,

A is diagonalizable.

5.3.35

A factorization

A=PDP−1

is not unique. For

A=

	9	2
	−12	−1

,

one factorization is

P=

	1	1
	−2	−3

,

D=

	5	0
	0	3

,

and

P−1=

	3	1
	−2	−1

.

Use this information with

D1=

	3	0
	0	5

to find a matrix

P1

such that

A=P1D1P−11.

P1=

	1	1
	−3	−2

(Type an integer or simplified fraction for each matrix element.)

5.3.35

A factorization

A=PDP−1

is not unique. For

A=

	6	1
	−6	1

,

one factorization is

P=

	1	1
	−2	−3

,

D=

	4	0
	0	3

,

and

P−1=

	3	1
	−2	−1

.

Use this information with

D1=

	3	0
	0	4

to find a matrix

P1

such that

A=P1D1P−11.

P1=

	1	1
	−3	−2

(Type an integer or simplified fraction for each matrix element.)

5.3.35

A factorization

A=PDP−1

is not unique. For

A=

	9	2
	−12	−1

,

one factorization is

P=

	1	1
	−2	−3

,

D=

	5	0
	0	3

,

and

P−1=

	3	1
	−2	−1

.

Use this information with

D1=

	3	0
	0	5

to find a matrix

P1

such that

A=P1D1P−11.

P1=

	1	1
	−3	−2

(Type an integer or simplified fraction for each matrix element.)

5.3.37

Identify a nonzero

2×2

matrix that is invertible but not diagonalizable.

Choose the correct answer below.

	1	1
	0	1

	1	0
	0	0

	1	1
	1	1

	1	0
	1	0

	1	1
	0	0

	1	0
	0	1

5.3.37

Identify a nonzero

2×2

matrix that is invertible but not diagonalizable.

Choose the correct answer below.

	1	1
	0	0

	1	1
	1	1

	1	0
	0	1

	1	1
	0	1

	1	0
	0	0

	1	0
	1	0

5.3.37

Identify a nonzero

2×2

matrix that is invertible but not diagonalizable.

Choose the correct answer below.

	1	0
	0	1

	1	0
	1	0

	1	1
	1	1

	1	1
	0	0

	1	1
	0	1

	1	0
	0	0

5.5.1

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	1	−17
	1	9

The eigenvalues of

	1	−17
	1	9

are

5±i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

Find a basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0.

Choose the correct answer below.

	−4+i
	1

	i
	1

	1
	−4+i

	−4−i
	1

	1
	−4−i

	−i
	1

Find a basis for the eigenspace corresponding to the eigenvalue

a−bi,

where

b>0.

Choose the correct answer below.

	−4+i
	1

	−i
	1

	−4−i
	1

	i
	1

	1
	−4−i

	1
	−4+i

5.5.1

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	1	−17
	1	9

The eigenvalues of

	1	−17
	1	9

are

5±i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

Find a basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0.

Choose the correct answer below.

	−i
	1

	1
	−4−i

	−4+i
	1

	1
	−4+i

	−4−i
	1

	i
	1

Find a basis for the eigenspace corresponding to the eigenvalue

a−bi,

where

b>0.

Choose the correct answer below.

	−i
	1

	1
	−4−i

	i
	1

	1
	−4+i

	−4+i
	1

	−4−i
	1

5.5.1

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	1	−26
	1	11

The eigenvalues of

	1	−26
	1	11

are

6±i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

Find a basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0.

Choose the correct answer below.

	−i
	1

	−5−i
	1

	−5+i
	1

	1
	−5+i

	1
	−5−i

	i
	1

Find a basis for the eigenspace corresponding to the eigenvalue

a−bi,

where

b>0.

Choose the correct answer below.

	1
	−5+i

	−5−i
	1

	1
	−5−i

	i
	1

	−5+i
	1

	−i
	1

5.5.3

Let the matrix

	2	1
	−45	8

act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

Select all that apply.

λ=5+6i;

v=

	3−6i
	45

λ=−5+6i;

v=

	3−6i
	6

λ=5−6i;

v=

	3−6i
	6

λ=5−6i;

v=

	3+6i
	45

λ=6+5i;

v=

	3−6i
	45

λ=5+6i;

v=

	3+6i
	5

λ=6−5i;

v=

	3+6i
	45

λ=−5−6i;

v=

	3+6i
	5

5.5.3

Let the matrix

	2	1
	−25	8

act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

Select all that apply.

λ=5−4i;

v=

	3−4i
	4

λ=5+4i;

v=

	3+4i
	5

λ=4+5i;

v=

	3−4i
	25

λ=5+4i;

v=

	3−4i
	25

λ=5−4i;

v=

	3+4i
	25

λ=4−5i;

v=

	3+4i
	25

λ=−5+4i;

v=

	3−4i
	4

λ=−5−4i;

v=

	3+4i
	5

5.5.3

Let the matrix

	6	1
	−17	4

act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

Select all that apply.

λ=5+4i;

v=

	−1+4i
	5

λ=−5−4i;

v=

	−1+4i
	5

λ=5−4i;

v=

	−1−4i
	4

λ=4+5i;

v=

	−1−4i
	17

λ=4−5i;

v=

	−1+4i
	17

λ=5+4i;

v=

	−1−4i
	17

λ=5−4i;

v=

	−1+4i
	17

λ=−5+4i;

v=

	−1−4i
	4

5.5.5

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	0	1
	−162	18

The eigenvalues of

	0	1
	−162	18

are

9±9i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

Find a basis for the eigenspace corresponding to the eigenvalue

a+bi.

Choose the correct answer below.

	1
	9+9i

	1
	9+i

	9+i
	1

	9i
	1

	1
	9i

	9+9i
	1

Find a basis for the eigenspace corresponding to the eigenvalue

a−bi.

Choose the correct answer below.

	1
	9−i

	9−i
	1

	−9i
	1

	9−9i
	1

	1
	9−9i

	1
	−9i

5.5.6

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	2	−2
	2	2

The eigenvalues of

	2	−2
	2	2

are

2±2i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0,

is

	i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a−bi

where

b>0,

is

	−i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

5.5.6

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	5	4
	−4	5

The eigenvalues of

	5	4
	−4	5

are

5±4i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0,

is

	−i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a−bi

where

b>0,

is

	i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

5.5.6

Let the matrix below act on

ℂ2.

Find the eigenvalues and a basis for each eigenspace in

ℂ2.

	1	3
	−3	1

The eigenvalues of

	1	3
	−3	1

are

1±3i.

(Type an exact answer, using radicals and

i

as needed. Use a comma to separate answers as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a+bi,

where

b>0,

is

	−i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

A basis for the eigenspace corresponding to the eigenvalue

a−bi

where

b>0,

is

	i
	1

.

(Type an exact answer, using radicals and

i

as needed.)

5.5.7

List the eigenvalues of

A.

The transformation

x↦Ax

is the composition of a rotation and a scaling. Give the angle

φ

of the rotation, where

−π<φ≤π,

and give the scale factor r.

A=

43	−4
4	43

The eigenvalues of A are

λ=43±4i.

(Simplify your answer. Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=π6

(Simplify your answer. Type an exact answer, using

π

as needed.)

r=8

(Simplify your answer.)

5.5.7

List the eigenvalues of

A.

The transformation

x↦Ax

is the composition of a rotation and a scaling. Give the angle

φ

of the rotation, where

−π<φ≤π,

and give the scale factor r.

A=

−3	1
−1	−3

The eigenvalues of A are

λ=−3±i.

(Simplify your answer. Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=−5π6

(Simplify your answer. Type an exact answer, using

π

as needed.)

r=2

(Simplify your answer.)

5.5.7

List the eigenvalues of

A.

The transformation

x↦Ax

is the composition of a rotation and a scaling. Give the angle

φ

of the rotation, where

−π<φ≤π,

and give the scale factor r.

A=

33	−3
3	33

The eigenvalues of A are

λ=33±3i.

(Simplify your answer. Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=π6

(Simplify your answer. Type an exact answer, using

π

as needed.)

r=6

(Simplify your answer.)

5.5.9

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −33 −13 13 −33

The eigenvalues of A are

λ=−33±13i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=5π6

(Type an exact answer, using

π

as needed.)

r=23

(Type an exact answer, using radicals as needed.)

5.5.9

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −34 14 −14 −34

The eigenvalues of A are

λ=−34±14i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=−5π6

(Type an exact answer, using

π

as needed.)

r=12

(Type an exact answer, using radicals as needed.)

5.5.9

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −37 17 −17 −37

The eigenvalues of A are

λ=−37±17i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=−5π6

(Type an exact answer, using

π

as needed.)

r=27

(Type an exact answer, using radicals as needed.)

5.5.10

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= 2 −2 2 2

The eigenvalues of A are

λ=2±2i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=π4

(Type an exact answer, using

π

as needed.)

r=22

(Type an exact answer, using radicals as needed.)

5.5.10

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −4 −4 4 −4

The eigenvalues of A are

λ=−4±4i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=3π4

(Type an exact answer, using

π

as needed.)

r=42

(Type an exact answer, using radicals as needed.)

5.5.10

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −6 6 −6 −6

The eigenvalues of A are

λ=−6±6i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=−3π4

(Type an exact answer, using

π

as needed.)

r=122

(Type an exact answer, using radicals as needed.)

5.5.11

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= −0.8 0.8 −0.8 −0.8

The eigenvalues of A are

λ=−45±45i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=−3π4

(Type an exact answer, using

π

as needed.)

r=425

(Type an exact answer, using radicals as needed.)

5.5.12

List the eigenvalues of A. The transformation x↦Ax is the composition of a rotation and a scaling. Give the angle φ of the rotation, where −π<φ≤π, and give the scale factor r.

A= 0 −0.8 0.8 0

The eigenvalues of A are

λ=±45i.

(Use a comma to separate answers as needed. Type an exact answer, using radicals and

i

as needed.)

φ=π2

(Simplify your answer. Type an exact answer, using

π

as needed.)

r=45

(Simplify your answer. Type an exact answer, using radicals as needed.)

5.5.13

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1	−5
	1	5

has the form

A=PCP−1.

The eigenvalues of A are

3−i

and

3+i.

The corresponding eigenvectors are

v1=

	−2−i
	1

and

v2=

	−2+i
	1

,

respectively.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	−2	−1
	1	0

,

	3	−1
	1	3

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.13

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1	−10
	1	7

has the form

A=PCP−1.

The eigenvalues of A are

4−i

and

4+i.

The corresponding eigenvectors are

v1=

	−3−i
	1

and

v2=

	−3+i
	1

,

respectively.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	−3	−1
	1	0

,

	4	−1
	1	4

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.13

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1	−2
	1	3

has the form

A=PCP−1.

The eigenvalues of A are

2−i

and

2+i.

The corresponding eigenvectors are

v1=

	−1−i
	1

and

v2=

	−1+i
	1

,

respectively.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	−1	−1
	1	0

,

	2	−1
	1	2

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.16

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	2	−37
	1	4

has the form

A=PCP−1.

The eigenvalues of A are

3+6i

and

3−6i

with corresponding eigenvectors

	6+i
	−i

and

	6−i
	i

.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	6	1
	0	−1

,

	3	6
	−6	3

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.16

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	5	−5
	1	7

has the form

A=PCP−1.

The eigenvalues of A are

6+2i

and

6−2i

with corresponding eigenvectors

	2+i
	−i

and

	2−i
	i

.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	2	1
	0	−1

,

	6	2
	−2	6

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.16

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	4	−5
	1	2

has the form

A=PCP−1.

The eigenvalues of A are

3+2i

and

3−2i

with corresponding eigenvectors

	2−i
	−i

and

	2+i
	i

.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	2	−1
	0	−1

,

	3	2
	−2	3

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.17

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1	−0.75
	2.4	−1.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	12	14
	1	0

,

	−15	35
	−35	−15

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.17

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1	−2
	0.8	−1.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	32	12
	1	0

,

	−15	25
	−25	−15

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.17

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	0.8	−0.6
	3	−1.6

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	25	15
	1	0

,

	−25	35
	−35	−25

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.18

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	0.6	−0.08
	1	0.2

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	15	15
	1	0

,

	25	15
	−15	25

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.18

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1.2	−0.8
	1.6	−0.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	12	12
	1	0

,

	25	45
	−45	25

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.18

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that

A=

	1.2	−0.8
	1.6	−0.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	12	12
	1	0

,

	25	45
	−45	25

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.19

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	1.52	−0.7
	0.56	0.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	1	12
	1	0

,

	2425	725
	−725	2425

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.19

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	1.2	−0.5
	1.04	0.4

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	513	1526
	1	0

,

	45	35
	−35	45

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

5.5.19

Find an invertible matrix P and a matrix C of the form

	a	−b
	b	a

such that the matrix

A=

	1	−0.5
	0.8	0.6

has the form

A=PCP−1.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

The matrices P and C are

	14	34
	1	0

,

	45	35
	−35	45

.

(Use a comma to separate answers as needed.)

There is no matrix C of the form

	a	−b
	b	a

and no invertible matrix P such that

A=PCP−1.

6.1.1

Compute

u•u,

v•u,

and

v•uu•u

using the vectors

u=

	−2
	5

and

v=

	7
	6

.

u•u=29

(Simplify your answer.)

v•u=16

(Simplify your answer.)

v•uu•u=1629

(Type an integer or a simplified fraction.)

6.1.1

Compute

u•u,

v•u,

and

v•uu•u

using the vectors

u=

	−1
	4

and

v=

	5
	3

.

u•u=17

(Simplify your answer.)

v•u=7

(Simplify your answer.)

v•uu•u=717

(Type an integer or a simplified fraction.)

6.1.1

Compute

u•u,

v•u,

and

v•uu•u

using the vectors

u=

	−2
	3

and

v=

	6
	7

.

u•u=13

(Simplify your answer.)

v•u=9

(Simplify your answer.)

v•uu•u=913

(Type an integer or a simplified fraction.)

6.1.2

Compute

w•w,

x•w,

and

x•ww•w

using the vectors

w=

	3
	−4
	−5

and

x=

	6
	−3
	3

.

w•w=50

(Simplify your answer. Type an integer or a simplified fraction.)

x•w=15

(Simplify your answer. Type an integer or a simplified fraction.)

x•ww•w=310

(Simplify your answer. Type an integer or a simplified fraction.)

6.1.2

Compute

w•w,

x•w,

and

x•ww•w

using the vectors

w=

	3
	−4
	−5

and

x=

	7
	−3
	2

.

w•w=50

(Simplify your answer. Type an integer or a simplified fraction.)

x•w=23

(Simplify your answer. Type an integer or a simplified fraction.)

x•ww•w=2350

(Simplify your answer. Type an integer or a simplified fraction.)

6.1.4

Compute

1u•uu,

where

u=

	−4
	5

.

1u•uu=

	−441
	541

(Type an integer or simplified fraction for each matrix element.)

6.1.4

Compute

1u•uu,

where

u=

	−2
	3

.

1u•uu=

	−213
	313

(Type an integer or simplified fraction for each matrix element.)

6.1.7

Compute

w

using

w=

	5
	1
	−5

.

w=51

(Type an exact answer, using radicals as needed.)

6.1.7

Compute

w

using

w=

	2
	2
	−7

.

w=57

(Type an exact answer, using radicals as needed.)

6.1.9

Find a unit vector in the direction of the given vector.

	72
	−320

A unit vector in the direction of the given vector is

	941
	−4041

.

(Type an exact answer, using radicals as needed.)

6.1.9

Find a unit vector in the direction of the given vector.

	77
	−264

A unit vector in the direction of the given vector is

	77275
	−264275

.

(Type an exact answer, using radicals as needed.)

6.1.10

Find a unit vector in the direction of the given vector.

	−2
	−18
	−2

A unit vector in the direction of the given vector is

	−183
	−983
	−183

.

(Type an exact answer, using radicals as needed.)

6.1.10

Find a unit vector in the direction of the given vector.

	−3
	−18
	3

A unit vector in the direction of the given vector is

	−138
	−638
	138

.

(Type an exact answer, using radicals as needed.)

6.1.13

Find the distance between

x=

	12
	−3

and

y=

	−4
	−9

.

The distance between x and y is

273.

(Type an exact answer, using radicals as needed.)

6.1.13

Find the distance between

x=

	10
	−4

and

y=

	−1
	−9

.

The distance between x and y is

146.

(Type an exact answer, using radicals as needed.)

6.1.13

Find the distance between

x=

	11
	−3

and

y=

	−1
	−9

.

The distance between x and y is

180.

(Type an exact answer, using radicals as needed.)

6.1.14

Find the distance between

u=

	0
	−6
	3

and

z=

	−2
	−1
	6

.

The distance between u and z is

38.

(Type an exact answer, using radicals as needed.)

6.1.14

Find the distance between

u=

	0
	−6
	2

and

z=

	−3
	−1
	5

.

The distance between u and z is

43.

(Type an exact answer, using radicals as needed.)

6.1.15

Determine if the following vectors are orthogonal.

a=

	8
	−3

,

b=

	−3
	−2

Select the correct choice below and fill in the answer box to complete your choice.

(Type an integer or a simplified fraction.)

The vectors a and b are not orthogonal because

a+b=.

The vectors a and b are not orthogonal because

a•b=−18.

The vectors a and b are orthogonal because

a•b=.

The vectors a and b are orthogonal because

a+b=.

6.1.15

Determine if the following vectors are orthogonal.

a=

	8
	−3

,

b=

	−2
	−2

Select the correct choice below and fill in the answer box to complete your choice.

(Type an integer or a simplified fraction.)

The vectors a and b are orthogonal because

a+b=.

The vectors a and b are not orthogonal because

a•b=−10.

The vectors a and b are orthogonal because

a•b=.

The vectors a and b are not orthogonal because

a+b=.

6.1.16

Determine if the following vectors are orthogonal.

u=

	10
	4
	2

,

v=

	1
	−4
	3

Select the correct choice below and fill in the answer box to complete your choice.

(Type an integer or a simplified fraction.)

The vectors u and v are orthogonal because

u•v=0.

The vectors u and v are not orthogonal because

u•v=.

The vectors u and v are not orthogonal because

u+v=.

The vectors u and v are orthogonal because

u+v=.

6.1.16

Determine if the following vectors are orthogonal.

u=

	10
	5
	15

,

v=

	1
	−5
	1

Select the correct choice below and fill in the answer box to complete your choice.

(Type an integer or a simplified fraction.)

The vectors u and v are orthogonal because

u•v=0.

The vectors u and v are orthogonal because

u+v=.

The vectors u and v are not orthogonal because

u+v=.

The vectors u and v are not orthogonal because

u•v=.

6.1.18

Determine if the following vectors are orthogonal.

u=

−4

4

9

0

,

v=

1

−8

16

−4

Are the two vectors orthogonal?

(Type an integer or a fraction.)

The vectors u and v

are

orthogonal because

u+v=.

The vectors u and v

are

orthogonal because

u•v=.

The vectors u and v

are not

orthogonal because

u•v=108.

The vectors u and v

are not

orthogonal because

u+v=.

6.1.19

Determine whether the statement below is true or false. Justify the answer. The vector is in

ℝn.

v•v=v2

Choose the correct answer below.

The statement is false. By the definition of the length of a vector

v,

v=v•v.

It follows that

v2=(v•v)2.

The statement is true. By the definition of the length of a vector

v,

v=v•v.

It follows that

v2=(v•v)2.

The statement is false. The expression

v•v

represents a vector, whereas

v2

is a scalar.

The statement is true. By the definition of the length of a vector

v,

v=v•v.

6.1.19

Determine whether the statement below is true or false. Justify the answer. The vector is in

ℝn.

v•v=v2

Choose the correct answer below.

The statement is true. By the definition of the length of a vector

v,

v=v•v.

The statement is false. The expression

v•v

represents a vector, whereas

v2

is a scalar.

The statement is false. By the definition of the length of a vector

v,

v=v•v.

It follows that

v2=(v•v)2.

The statement is true. By the definition of the length of a vector

v,

v=v•v.

It follows that

v2=(v•v)2.

6.1.22

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

u2+v2=u+v2,

then

u

and

v

are orthogonal.

Choose the correct answer below.

The statement is false. Two vectors

u

and

v

are orthogonal if

u•v=0.

If

u2+v2=u+v2,

then

u•v=1.

The statement is false. If

u2+v2=u+v2,

then

u

and

v

are orthogonal complements.

The statement is true. By the Pythagorean Theorem, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

The statement is true. By the definition of the inner product, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

6.1.22

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

u2+v2=u+v2,

then

u

and

v

are orthogonal.

Choose the correct answer below.

The statement is true. By the Pythagorean Theorem, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

The statement is true. By the definition of the inner product, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

The statement is false. Two vectors

u

and

v

are orthogonal if

u•v=0.

If

u2+v2=u+v2,

then

u•v=1.

The statement is false. If

u2+v2=u+v2,

then

u

and

v

are orthogonal complements.

6.1.22

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

u2+v2=u+v2,

then

u

and

v

are orthogonal.

Choose the correct answer below.

The statement is true. By the definition of the inner product, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

The statement is true. By the Pythagorean Theorem, two vectors

u

and

v

are orthogonal if and only if

u+v2=u2+v2.

The statement is false. If

u2+v2=u+v2,

then

u

and

v

are orthogonal complements.

The statement is false. Two vectors

u

and

v

are orthogonal if

u•v=0.

If

u2+v2=u+v2,

then

u•v=1.

6.1.23

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If vectors

v1,

...,

vp

span a subspace W and if

x

is orthogonal to each

vj

for

j=1,

..., p, then

x

is in

W⊥.

Choose the correct answer below.

The statement is true. A vector

x

is in

W⊥

if and only if

dist(x,v)=0

for any vector

v

in a set V that spans W.

The statement is true. If

x

is orthogonal to each

vj,

then

x

is also orthogonal to any linear combination of those

vj.

Since any vector in W can be described as a linear combination of

vj,

x

is orthogonal to all vectors in W.

The statement is false. It is known that a vector

x

is in

W⊥

if and only if

x

is not orthogonal to every vector in a set that spans W. The vectors

v1,

...,

vp

make up a set V that spans the subspace W. This means

x

is orthogonal to every vector in V and, therefore,

x

is not in

W⊥.

The statement is false. It is known that a vector

x

is in

W⊥

if and only if

x

is orthogonal to every vector

w

in W. The vectors

v1,

...,

vp

make up a set V that spans the subspace W, but this is not enough to say that

x

is in

W⊥.

6.1.23

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If vectors

v1,

...,

vp

span a subspace W and if

x

is orthogonal to each

vj

for

j=1,

..., p, then

x

is in

W⊥.

Choose the correct answer below.

The statement is true. A vector

x

is in

W⊥

if and only if

dist(x,v)=0

for any vector

v

in a set V that spans W.

The statement is false. It is known that a vector

x

is in

W⊥

if and only if

x

is not orthogonal to every vector in a set that spans W. The vectors

v1,

...,

vp

make up a set V that spans the subspace W. This means

x

is orthogonal to every vector in V and, therefore,

x

is not in

W⊥.

The statement is false. It is known that a vector

x

is in

W⊥

if and only if

x

is orthogonal to every vector

w

in W. The vectors

v1,

...,

vp

make up a set V that spans the subspace W, but this is not enough to say that

x

is in

W⊥.

The statement is true. If

x

is orthogonal to each

vj,

then

x

is also orthogonal to any linear combination of those

vj.

Since any vector in W can be described as a linear combination of

vj,

x

is orthogonal to all vectors in W.

6.1.24

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

x

is orthogonal to every vector in a subspace W, then

x

is in

W⊥.

Choose the correct answer below.

The statement is false. If

x

is orthogonal to every vector in a subspace W, then

x

is in W and so

x

cannot be in

W⊥.

The statement is true. If

x

is orthogonal to every vector in a subspace W, then

x=0.

The zero vector is in every subspace, so

x

must be in

W⊥.

The statement is false. A vector

x

is in

W⊥

if and only if

x

is orthogonal to every vector in a set that spans W.

The statement is true. If

x

is orthogonal to every vector in W, then

x

is said to be orthogonal to W. The set of all vectors

x

that are orthogonal to W is denoted

W⊥.

6.1.24

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

x

is orthogonal to every vector in a subspace W, then

x

is in

W⊥.

Choose the correct answer below.

The statement is true. If

x

is orthogonal to every vector in W, then

x

is said to be orthogonal to W. The set of all vectors

x

that are orthogonal to W is denoted

W⊥.

The statement is false. If

x

is orthogonal to every vector in a subspace W, then

x

is in W and so

x

cannot be in

W⊥.

The statement is true. If

x

is orthogonal to every vector in a subspace W, then

x=0.

The zero vector is in every subspace, so

x

must be in

W⊥.

The statement is false. A vector

x

is in

W⊥

if and only if

x

is orthogonal to every vector in a set that spans W.

6.1.24

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

If

x

is orthogonal to every vector in a subspace W, then

x

is in

W⊥.

Choose the correct answer below.

The statement is false. A vector

x

is in

W⊥

if and only if

x

is orthogonal to every vector in a set that spans W.

The statement is true. If

x

is orthogonal to every vector in W, then

x

is said to be orthogonal to W. The set of all vectors

x

that are orthogonal to W is denoted

W⊥.

The statement is true. If

x

is orthogonal to every vector in a subspace W, then

x=0.

The zero vector is in every subspace, so

x

must be in

W⊥.

The statement is false. If

x

is orthogonal to every vector in a subspace W, then

x

is in W and so

x

cannot be in

W⊥.

6.1.27

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

For a square matrix A, vectors in

Col A

are orthogonal to vectors in

Nul A.

Choose the correct answer below.

The statement is false. By the theorem of orthogonal complements,

(Row A)⊥=Nul A.

This means that vectors in

Row A

are orthogonal to vectors in

Nul A,

so vectors in

Col A

cannot also be orthogonal to vectors in

Nul A.

The statement is false. By the theorem of orthogonal complements, it is known that vectors in

Col A

are orthogonal to vectors in

Nul AT.

Using the definition of orthogonal complements, vectors in

Col A

are orthogonal to vectors in

Nul A

if and only if the rows and columns of A are the same, which is not necessarily true.

The statement is true. By the theorem of orthogonal complements, it is known that vectors in

Col A

are orthogonal to vectors in

Nul AT.

Using the definition of orthogonal complements, vectors in

Col A

are orthogonal to vectors in

Nul A

if and only if the rows and columns of A are the same, which is true for any square matrix.

The statement is true. By the theorem of orthogonal complements,

(Row A)⊥=Nul A.

By the definition of orthogonal complements, it follows that vectors in

Col A

are orthogonal to vectors in

Nul A.

6.1.27

Determine whether the statement below is true or false. Justify the answer. The vectors are in

ℝn.

For a square matrix A, vectors in

Col A

are orthogonal to vectors in

Nul A.

Choose the correct answer below.

The statement is false. By the theorem of orthogonal complements, it is known that vectors in

Col A

are orthogonal to vectors in

Nul AT.

Using the definition of orthogonal complements, vectors in

Col A

are orthogonal to vectors in

Nul A

if and only if the rows and columns of A are the same, which is not necessarily true.

The statement is false. By the theorem of orthogonal complements,

(Row A)⊥=Nul A.

This means that vectors in

Row A

are orthogonal to vectors in

Nul A,

so vectors in

Col A

cannot also be orthogonal to vectors in

Nul A.

The statement is true. By the theorem of orthogonal complements,

(Row A)⊥=Nul A.

By the definition of orthogonal complements, it follows that vectors in

Col A

are orthogonal to vectors in

Nul A.

The statement is true. By the theorem of orthogonal complements, it is known that vectors in

Col A

are orthogonal to vectors in

Nul AT.

Using the definition of orthogonal complements, vectors in

Col A

are orthogonal to vectors in

Nul A

if and only if the rows and columns of A are the same, which is true for any square matrix.

6.2.2

Determine whether the set of vectors is orthogonal.

−2

1

1

,

1

2

0

,

−2

1

−5

Is the set of vectors orthogonal?

The set of vectors is orthogonal because each pair of distinct vectors from the set is orthogonal.

The set of vectors is not orthogonal because at least one pair of distinct vectors from the set is not orthogonal.

The set of vectors is orthogonal because

−2

1

1

is orthogonal to

1

2

0

and

1

2

0

is orthogonal to

−2

1

−5

.

The set of vectors is orthogonal because the vectors are linearly independent.

6.2.2

Determine whether the set of vectors is orthogonal.

−4

1

1

,

1

4

0

,

−4

1

−17

Is the set of vectors orthogonal?

The set of vectors is orthogonal because

−4

1

1

is orthogonal to

1

4

0

and

1

4

0

is orthogonal to

−4

1

−17

.

The set of vectors is orthogonal because the vectors are linearly independent.

The set of vectors is orthogonal because each pair of distinct vectors from the set is orthogonal.

The set of vectors is not orthogonal because at least one pair of distinct vectors from the set is not orthogonal.

6.2.2

Determine whether the set of vectors is orthogonal.

−2

1

1

,

1

2

0

,

−2

1

−5

Is the set of vectors orthogonal?

The set of vectors is orthogonal because each pair of distinct vectors from the set is orthogonal.

The set of vectors is orthogonal because

−2

1

1

is orthogonal to

1

2

0

and

1

2

0

is orthogonal to

−2

1

−5

.

The set of vectors is orthogonal because the vectors are linearly independent.

The set of vectors is not orthogonal because at least one pair of distinct vectors from the set is not orthogonal.

6.2.3

Determine whether the set of vectors is orthogonal.

	2
	−5
	−1

,

	−8
	−4
	4

,

	3
	1
	1

Is the set of vectors orthogonal?

No

Yes

6.2.3

Determine whether the set of vectors is orthogonal.

	1
	3
	−1

,

	0
	−1
	−3

,

	−10
	3
	−1

Is the set of vectors orthogonal?

Yes

No

6.2.4

Determine whether the set of vectors is orthogonal.

4

−5

−7

,

0

0

0

,

2

−4

4

Is the set of vectors orthogonal?

Yes

No

6.2.4

Determine whether the set of vectors is orthogonal.

0

0

0

,

2

−3

−7

,

4

−2

2

Is the set of vectors orthogonal?

No

Yes

6.2.7

Show that

u1, u2

is an orthogonal basis for

ℝ2.

Then express x as a linear combination of the

u's.

u1=

	2
	−4

,

u2=

	8
	4

,

and

x=

	6
	−7

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for

ℝ2?

Select all that apply.

The vectors must form an orthogonal set.

The vectors must all have a length of 1.

The vectors must span

ℝ2.

The distance between any pair of distinct vectors must be constant.

Which theorem could help prove one of these criteria from another?

If

S=u1, ..., up

is an orthogonal set of nonzero vectors in

ℝn,

then S is linearly independent and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

is a basis in

ℝp,

then the members of S span

ℝp

and hence form an orthogonal set.

If

S=u1, ..., up

and each

ui

has length 1, then S is an orthogonal set and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set.

What calculation shows that

u1, u2

is an orthogonal basis for

ℝ2?

Since the

inner product of u1 and u2

is

0,

the vectors

form an orthogonal set.

From the theorem above, this proves that the vectors are also

a basis for ℝ2

because they are two

linearly independent vectors in ℝ2.

Express x as a linear combination of the

u's.

x=2u1+14u2

(Simplify your answers. Use integers or fractions for any numbers in the equation.)

6.2.7

Show that

u1, u2

is an orthogonal basis for

ℝ2.

Then express x as a linear combination of the

u's.

u1=

	2
	−6

,

u2=

	12
	4

,

and

x=

	2
	−2

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for

ℝ2?

Select all that apply.

The vectors must span

ℝ2.

The vectors must form an orthogonal set.

The distance between any pair of distinct vectors must be constant.

The vectors must all have a length of 1.

Which theorem could help prove one of these criteria from another?

If

S=u1, ..., up

and each

ui

has length 1, then S is an orthogonal set and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

is a basis in

ℝp,

then the members of S span

ℝp

and hence form an orthogonal set.

If

S=u1, ..., up

and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set.

If

S=u1, ..., up

is an orthogonal set of nonzero vectors in

ℝn,

then S is linearly independent and hence is a basis for the subspace spanned by S.

What calculation shows that

u1, u2

is an orthogonal basis for

ℝ2?

Since the

inner product of u1 and u2

is

0,

the vectors

form an orthogonal set.

From the theorem above, this proves that the vectors are also

a basis for ℝ2

because they are two

linearly independent vectors in ℝ2.

Express x as a linear combination of the

u's.

x=25u1+110u2

(Simplify your answers. Use integers or fractions for any numbers in the equation.)

6.2.10

Show that

u1, u2, u3

is an orthogonal basis for

ℝ3.

Then express x as a linear combination of the

u's.

u1=

	4
	−4
	0

,

u2=

	2
	2
	−1

,

u3=

	1
	1
	4

,

and

x=

	4
	−3
	1

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for a subspace W of

ℝn?

Select all that apply.

The distance between any pair of distinct vectors must be constant.

The vectors must all have a length of 1.

The vectors must form an orthogonal set.

The vectors must span W.

Which theorem could help prove one of these criteria from another?

If

S=u1, ..., up

and each

ui

has length 1, then S is an orthogonal set and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

is a basis in

ℝp,

then the members of S span

ℝp

and hence form an orthogonal set.

If

S=u1, ..., up

is an orthogonal set of nonzero vectors in

ℝn,

then S is linearly independent and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set.

Which calculations should be performed next?

(Simplify your answers.)

u1−u2=

u1−u3=

u2−u3=

u1•u2=0

u1•u3=0

u2•u3=0

u1•u1=

u2•u2=

u3•u3=

How do these calculations show that

u1, u2, u3

is an orthogonal basis for

ℝ3?

Since each

inner product

is

0,

the vectors

form an orthogonal set.

From the theorem above, this proves that the vectors are also

a basis.

Express x as a linear combination of the

u's.

x=78u1+19u2+518u3

(Use integers or fractions for any numbers in the equation.)

6.2.10

Show that

u1, u2, u3

is an orthogonal basis for

ℝ3.

Then express x as a linear combination of the

u's.

u1=

	3
	−3
	0

,

u2=

	2
	2
	−1

,

u3=

	1
	1
	4

,

and

x=

	5
	−2
	1

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for a subspace W of

ℝn?

Select all that apply.

The distance between any pair of distinct vectors must be constant.

The vectors must all have a length of 1.

The vectors must span W.

The vectors must form an orthogonal set.

Which theorem could help prove one of these criteria from another?

If

S=u1, ..., up

and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set.

If

S=u1, ..., up

is an orthogonal set of nonzero vectors in

ℝn,

then S is linearly independent and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

is a basis in

ℝp,

then the members of S span

ℝp

and hence form an orthogonal set.

If

S=u1, ..., up

and each

ui

has length 1, then S is an orthogonal set and hence is a basis for the subspace spanned by S.

Which calculations should be performed next?

(Simplify your answers.)

u1•u2=0

u1•u3=0

u2•u3=0

u1•u1=

u2•u2=

u3•u3=

u1−u2=

u1−u3=

u2−u3=

How do these calculations show that

u1, u2, u3

is an orthogonal basis for

ℝ3?

Since each

inner product

is

0,

the vectors

form an orthogonal set.

From the theorem above, this proves that the vectors are also

a basis.

Express x as a linear combination of the

u's.

x=76u1+59u2+718u3

(Use integers or fractions for any numbers in the equation.)

6.2.10

Show that

u1, u2, u3

is an orthogonal basis for

ℝ3.

Then express x as a linear combination of the

u's.

u1=

	4
	−4
	0

,

u2=

	3
	3
	−1

,

u3=

	1
	1
	6

,

and

x=

	4
	−3
	1

Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for a subspace W of

ℝn?

Select all that apply.

The distance between any pair of distinct vectors must be constant.

The vectors must form an orthogonal set.

The vectors must all have a length of 1.

The vectors must span W.

Which theorem could help prove one of these criteria from another?

If

S=u1, ..., up

and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set.

If

S=u1, ..., up

is an orthogonal set of nonzero vectors in

ℝn,

then S is linearly independent and hence is a basis for the subspace spanned by S.

If

S=u1, ..., up

is a basis in

ℝp,

then the members of S span

ℝp

and hence form an orthogonal set.

If

S=u1, ..., up

and each

ui

has length 1, then S is an orthogonal set and hence is a basis for the subspace spanned by S.

Which calculations should be performed next?

(Simplify your answers.)

u1•u1=

u2•u2=

u3•u3=

u1•u2=0

u1•u3=0

u2•u3=0

u1−u2=

u1−u3=

u2−u3=

How do these calculations show that

u1, u2, u3

is an orthogonal basis for

ℝ3?

Since each

inner product

is

0,

the vectors

form an orthogonal set.

From the theorem above, this proves that the vectors are also

a basis.

Express x as a linear combination of the

u's.

x=78u1+219u2+738u3

(Use integers or fractions for any numbers in the equation.)

6.2.11

Compute the orthogonal projection of

	1
	7

onto the line through

	−3
	9

and the origin.

The orthogonal projection is

	−2
	6

.

(Simplify your answer.)

6.2.11

Compute the orthogonal projection of

	8
	9

onto the line through

	−3
	6

and the origin.

The orthogonal projection is

	−2
	4

.

(Simplify your answer.)

6.2.13

Let

y=

1

6

and

u=

5

−7

.

Write

y

as the sum of two orthogonal vectors, one in

Span {u}

and one orthogonal to

u.

y=y+z=

	−52
	72

+

	72
	52

(Type an integer or simplified fraction for each matrix element. List the terms in the same order as they appear in the original list.)

6.2.13

Let

y=

1

4

and

u=

3

−5

.

Write

y

as the sum of two orthogonal vectors, one in

Span {u}

and one orthogonal to

u.

y=y+z=

	−32
	52

+

	52
	32

(Type an integer or simplified fraction for each matrix element. List the terms in the same order as they appear in the original list.)

6.2.15

Let

y=

	4
	7

and

u=

	6
	8

.

Compute the distance from y to the line through u and the origin.

The distance from y to the line through u and the origin is

1.

(Simplify your answer.)

6.2.15

Let

y=

	9
	6

and

u=

	8
	6

.

Compute the distance from y to the line through u and the origin.

The distance from y to the line through u and the origin is

35.

(Simplify your answer.)

6.2.15

Let

y=

	4
	2

and

u=

	8
	6

.

Compute the distance from y to the line through u and the origin.

The distance from y to the line through u and the origin is

45.

(Simplify your answer.)

6.2.17

Determine whether the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.

u1=

	23
	−13
	13

 and u2=

	−13
	0
	23

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The set of vectors is orthogonal only. The normalized vectors for

u1

and

u2

are

	26
	−16
	16

and

	−15
	0
	25

,

respectively.

(Type exact answers, using radicals as needed.)  

The set of vectors is orthonormal.

The set of vectors is not orthogonal.

6.2.17

Determine whether the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.

u1=

	23
	−13
	13

 and u2=

	−15
	0
	25

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The set of vectors is orthogonal only. The normalized vectors for

u1

and

u2

are

	23
	−16
	16

and

	−55
	0
	255

,

respectively.

(Type exact answers, using radicals as needed.)  

The set of vectors is orthonormal.

The set of vectors is not orthogonal.

6.2.19

Determine if the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.

u=

	0.6
	−0.8

,

v=

	−0.8
	−0.6

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The set of vectors is orthogonal only. The normalized vectors for

u

and

v

are

and

,

respectively.

(Type an exact answer, using radicals as needed.)

The set of vectors is orthonormal.

The set of vectors is not orthogonal.

6.2.19

Determine if the set of vectors is orthonormal. If the set is only orthogonal, normalize the vectors to produce an orthonormal set.

u=

	−0.8
	0.6

,

v=

	0.6
	0.8

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

The set of vectors is orthogonal only. The normalized vectors for

u

and

v

are

and

,

respectively.

(Type an exact answer, using radicals as needed.)

The set of vectors is orthonormal.

The set of vectors is not orthogonal.

6.2.23

Determine whether the statement below is true or false. Justify the answer.

Not every linearly independent set in

ℝn

is an orthogonal set.

Choose the correct answer below.

The statement is true. For example, the vectors

	0
	1

and

	1
	1

are linearly independent but not orthogonal.

The statement is true. For example, the vectors

	1
	−1

and

	1
	1

are linearly independent but not orthogonal.

The statement is false. Every orthogonal set is linearly independent.

The statement is false. In every linearly independent set of two vectors in

ℝn,

one vector is a multiple of the other, so the vectors cannot be orthogonal.

6.2.23

Determine whether the statement below is true or false. Justify the answer.

Not every linearly independent set in

ℝn

is an orthogonal set.

Choose the correct answer below.

The statement is true. For example, the vectors

	0
	1

and

	1
	1

are linearly independent but not orthogonal.

The statement is false. In every linearly independent set of two vectors in

ℝn,

one vector is a multiple of the other, so the vectors cannot be orthogonal.

The statement is false. Every orthogonal set is linearly independent.

The statement is true. For example, the vectors

	1
	−1

and

	1
	1

are linearly independent but not orthogonal.

6.2.23

Determine whether the statement below is true or false. Justify the answer.

Not every linearly independent set in

ℝn

is an orthogonal set.

Choose the correct answer below.

The statement is false. In every linearly independent set of two vectors in

ℝn,

one vector is a multiple of the other, so the vectors cannot be orthogonal.

The statement is true. For example, the vectors

	0
	1

and

	1
	1

are linearly independent but not orthogonal.

The statement is false. Every orthogonal set is linearly independent.

The statement is true. For example, the vectors

	1
	−1

and

	1
	1

are linearly independent but not orthogonal.

6.2.24

Determine whether the statement below is true or false. Justify the answer.

Not every orthogonal set in

ℝn

is linearly independent.

Choose the correct answer below.

The statement is true. Every orthogonal set of nonzero vectors is linearly independent, but not every orthogonal set is linearly independent.

The statement is true. Orthogonal sets with fewer than n vectors in

ℝn

are not linearly independent.

The statement is false. Every orthogonal set of nonzero vectors is linearly independent and zero vectors cannot exist in orthogonal sets.

The statement is false. Orthogonal sets must be linearly independent in order to be orthogonal.

6.2.24

Determine whether the statement below is true or false. Justify the answer.

Not every orthogonal set in

ℝn

is linearly independent.

Choose the correct answer below.

The statement is false. Orthogonal sets must be linearly independent in order to be orthogonal.

The statement is true. Orthogonal sets with fewer than n vectors in

ℝn

are not linearly independent.

The statement is false. Every orthogonal set of nonzero vectors is linearly independent and zero vectors cannot exist in orthogonal sets.

The statement is true. Every orthogonal set of nonzero vectors is linearly independent, but not every orthogonal set is linearly independent.

6.2.25

Determine whether the statement below is true or false. Justify the answer. All vectors are in

ℝn.

If

y

is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.

Choose the correct answer below.

The statement is false. The weights in any linear combination can only be computed using row operations.

The statement is true. If the orthogonal set is

u1, ..., up,

then for each

y,

the weights in the linear combination

y=c1u1+⋯+cpup

can be computed by

cj=y•ujuj•uj,

where

j=1, ..., p.

The statement is false. The weights in a linear combination can only be computed without row operations on a matrix if one of the vectors is the zero vector.

The statement is true. If the orthogonal set is

u1, ..., up,

then for each

y,

the weights in the linear combination

y=c1u1+⋯+cpup

can be computed by

cj=y•ujy•y,

where

j=1, ..., p.

6.2.25

Determine whether the statement below is true or false. Justify the answer. All vectors are in

ℝn.

If

y

is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.

Choose the correct answer below.

The statement is false. The weights in any linear combination can only be computed using row operations.

The statement is true. If the orthogonal set is

u1, ..., up,

then for each

y,

the weights in the linear combination

y=c1u1+⋯+cpup

can be computed by

cj=y•ujy•y,

where

j=1, ..., p.

The statement is true. If the orthogonal set is

u1, ..., up,

then for each

y,

the weights in the linear combination

y=c1u1+⋯+cpup

can be computed by

cj=y•ujuj•uj,

where

j=1, ..., p.

The statement is false. The weights in a linear combination can only be computed without row operations on a matrix if one of the vectors is the zero vector.

6.2.26

Determine whether the statement below is true or false. Justify the answer. All vectors are in

ℝn.

If a set

S=u1, ..., up

has the property that

ui•uj=0

whenever

i≠j,

then S is an orthonormal set.

Choose the correct answer below.

The statement is true. To be orthonormal, the vectors in S must be orthogonal to each other.

The statement is false. To be orthonormal, S must be linearly independent of the subspace spanned by S.

The statement is true. Because

ui•uj=0

whenever

i≠j,

S is linearly independent and thus is an orthonormal set.

The statement is false. To be orthonormal, the vectors in S must be unit vectors as well as being orthogonal to each other.

6.2.26

Determine whether the statement below is true or false. Justify the answer. All vectors are in

ℝn.

If a set

S=u1, ..., up

has the property that

ui•uj=0

whenever

i≠j,

then S is an orthonormal set.

Choose the correct answer below.

The statement is true. To be orthonormal, the vectors in S must be orthogonal to each other.

The statement is true. Because

ui•uj=0

whenever

i≠j,

S is linearly independent and thus is an orthonormal set.

The statement is false. To be orthonormal, the vectors in S must be unit vectors as well as being orthogonal to each other.

The statement is false. To be orthonormal, S must be linearly independent of the subspace spanned by S.

6.2.28

Determine whether the statement below is true or false. Justify the answer. The vector is in

ℝn.

If the columns of an

m×n

matrix A are orthonormal, then the linear mapping

x↦Ax

preserves lengths.

Choose the correct answer below.

The statement is true. If A is a matrix with orthonormal columns, then

Ax=x.

The statement is false. Only nonzero orthonormal matrices preserve lengths.

The statement is true. If A is a matrix with orthonormal columns, then A must be a square matrix. This means the linear mapping

x↦Ax

must preserve lengths.

The statement is false. If A is a matrix with orthonormal columns, then the linear mapping

x↦Ax

would only preserve lengths if A is square.

6.2.28

Determine whether the statement below is true or false. Justify the answer. The vector is in

ℝn.

If the columns of an

m×n

matrix A are orthonormal, then the linear mapping

x↦Ax

preserves lengths.

Choose the correct answer below.

The statement is true. If A is a matrix with orthonormal columns, then

Ax=x.

The statement is false. If A is a matrix with orthonormal columns, then the linear mapping

x↦Ax

would only preserve lengths if A is square.

The statement is false. Only nonzero orthonormal matrices preserve lengths.

The statement is true. If A is a matrix with orthonormal columns, then A must be a square matrix. This means the linear mapping

x↦Ax

must preserve lengths.

6.2.29

Determine whether the statement below is true or false. Justify the answer.

A matrix with orthonormal columns is an orthogonal matrix.

Choose the correct answer below.

The statement is false. A matrix with orthonormal columns is an orthogonal matrix if the matrix is not square.

The statement is true. All matrices with orthonormal columns are orthogonal matrices.

The statement is true. All matrices with orthonormal rows and columns are orthogonal matrices.

The statement is false. A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.

6.2.29

Determine whether the statement below is true or false. Justify the answer.

A matrix with orthonormal columns is an orthogonal matrix.

Choose the correct answer below.

The statement is false. A matrix with orthonormal columns is an orthogonal matrix if the matrix is not square.

The statement is true. All matrices with orthonormal columns are orthogonal matrices.

The statement is false. A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.

The statement is true. All matrices with orthonormal rows and columns are orthogonal matrices.

6.2.32

Determine whether the statement below is true or false. Justify the answer.

An orthogonal matrix is invertible.

Choose the correct answer below.

The statement is true. An orthogonal matrix is a square invertible matrix U such that

U−1=UT.

The statement is false. Only nonzero orthogonal matrices are invertible.

The statement is true. An orthogonal matrix is a square invertible matrix U such that

U−1=U.

The statement is false. Only orthogonal matrices with orthonormal rows are invertible.

6.3.4

Verify that

u1,u2

is an orthogonal set, and then find the orthogonal projection of y onto

Spanu1,u2.

y=

4

5

−4

,

u1=

4

5

0

,

u2=

−5

4

0

To verify that

u1,u2

is an orthogonal set, find

u1•u2.

u1•u2=0

(Simplify your answer.)

The projection of y onto

Spanu1,u2

is

4

5

0

.

(Simplify your answers.)

6.3.4

Verify that

u1,u2

is an orthogonal set, and then find the orthogonal projection of y onto

Spanu1,u2.

y=

5

3

−2

,

u1=

4

5

0

,

u2=

−5

4

0

To verify that

u1,u2

is an orthogonal set, find

u1•u2.

u1•u2=0

(Simplify your answer.)

The projection of y onto

Spanu1,u2

is

5

3

0

.

(Simplify your answers.)

6.3.4

Verify that

u1,u2

is an orthogonal set, and then find the orthogonal projection of y onto

Spanu1,u2.

y=

3

5

−1

,

u1=

3

6

0

,

u2=

−6

3

0

To verify that

u1,u2

is an orthogonal set, find

u1•u2.

u1•u2=0

(Simplify your answer.)

The projection of y onto

Spanu1,u2

is

3

5

0

.

(Simplify your answers.)

6.3.8

Let W be the subspace spanned by

u1

and

u2,

and write

y

as the sum of a vector in W and a vector orthogonal to W.

y=

	−6
	4
	7

,

u1=

	1
	1
	2

,

u2=

	−3
	9
	−3

The sum is

y=y+z,

where

y=

	1
	5
	3

is in W and

z=

	−7
	−1
	4

is orthogonal to W.

(Simplify your answers.)

6.3.8

Let W be the subspace spanned by

u1

and

u2,

and write

y

as the sum of a vector in W and a vector orthogonal to W.

y=

	−2
	3
	7

,

u1=

	1
	1
	3

,

u2=

	−2
	5
	−1

The sum is

y=y+z,

where

y=

	65
	4
	285

is in W and

z=

	−165
	−1
	75

is orthogonal to W.

(Simplify your answers.)

6.3.8

Let W be the subspace spanned by

u1

and

u2,

and write

y

as the sum of a vector in W and a vector orthogonal to W.

y=

	−7
	2
	9

,

u1=

	1
	1
	3

,

u2=

	−2
	5
	−1

The sum is

y=y+z,

where

y=

	1
	92
	112

is in W and

z=

	−8
	−52
	72

is orthogonal to W.

(Simplify your answers.)

6.3.9

Let W be a subspace spanned by the

u's,

and write

y

as the sum of a vector in W and a vector orthogonal to W.

y=

	4
	2
	2
	−1

,

u1=

	1
	1
	0
	1

,

u2=

	−1
	5
	3
	−4

,

u3=

	−1
	0
	1
	1

y=

	4017
	5517
	−117
	−1017

+

	2817
	−2117
	3517
	−717

(Type an integer or simplified fraction for each matrix element.)

6.3.9

Let W be a subspace spanned by the

u's,

and write

y

as the sum of a vector in W and a vector orthogonal to W.

y=

	4
	6
	6
	−1

,

u1=

	1
	1
	0
	1

,

u2=

	−1
	5
	3
	−4

,

u3=

	−1
	0
	1
	1

y=

	8851
	13117
	16151
	−2251

+

	11651
	−2917
	14551
	−2951

(Type an integer or simplified fraction for each matrix element.)

6.3.12

Find the closest point to

y

in the subspace W spanned by

v1

and

v2.

y=

	3
	−1
	1
	19

,

v1=

	1
	−3
	−1
	1

,

v2=

	−1
	1
	0
	4

The closest point to

y

in W is the vector

	−2
	−2
	−2
	18

.

(Simplify your answer.)

6.3.12

Find the closest point to

y

in the subspace W spanned by

v1

and

v2.

y=

	7
	−1
	1
	14

,

v1=

	1
	−1
	−1
	2

,

v2=

	−2
	2
	0
	2

The closest point to

y

in W is the vector

	3
	−3
	−5
	12

.

(Simplify your answer.)

6.3.13

Find the best approximation to

z

by vectors of the form

c1v1+c2v2.

z=

	3
	−6
	2
	3

,

v1=

	2
	1
	2
	−1

,

v2=

	1
	−1
	0
	1

The best approximation to

z

is

	215
	−3910
	15
	3910

.

(Simplify your answer.)

6.3.13

Find the best approximation to

z

by vectors of the form

c1v1+c2v2.

z=

	3
	−6
	2
	3

,

v1=

	3
	−1
	2
	2

,

v2=

	1
	1
	0
	−1

The best approximation to

z

is

	136
	−6118
	259
	439

.

(Simplify your answer.)

6.3.13

Find the best approximation to

z

by vectors of the form

c1v1+c2v2.

z=

	4
	−5
	2
	4

,

v1=

	4
	2
	3
	−2

,

v2=

	1
	−1
	0
	1

The best approximation to

z

is

	5311
	−4511
	411
	4511

.

(Simplify your answer.)

6.3.15

Let

y=

6

−8

5

,

u1=

−2

−5

1

,

u2=

−3

1

−1

.

Find the distance from y to the plane in

ℝ3

spanned by

u1

and

u2.

The distance is

73110.

(Type an exact answer, using radicals as needed.)

6.3.15

Let

y=

5

−8

5

,

u1=

−3

−5

1

,

u2=

−4

2

−2

.

Find the distance from y to the plane in

ℝ3

spanned by

u1

and

u2.

The distance is

542.

(Type an exact answer, using radicals as needed.)

6.3.16

Let

y=

−3

−1

1

−13

,

v1=

1

2

−1

−2

,

and

v2=

4

1

0

3

.

Find the distance from y to the subspace W of

ℝ4

spanned by

v1

and

v2,

given that the closest point to y in W is

y=

−6

2

−2

−10

.

The distance is

6.

(Simplify your answer. Type an exact answer, using radicals as needed.)

6.3.16

Let

y=

−3

−1

1

−13

,

v1=

1

2

−1

−2

,

and

v2=

4

1

0

3

.

Find the distance from y to the subspace W of

ℝ4

spanned by

v1

and

v2,

given that the closest point to y in W is

y=

−6

2

−2

−10

.

The distance is

6.

(Simplify your answer. Type an exact answer, using radicals as needed.)

6.3.17

Let

y=

	3
	9
	1

,

u1=

	23
	−13
	−23

,

u2=

	23
	23
	13

,

and

W=Span u1,u2.

Complete parts (a) and (b).

a. Let

U=

	u1	u2

.

Compute

UTU

and

UUT.

UTU=

	1	0
	0	1

and

UUT=

	89	29	−29
	29	59	49
	−29	49	59

(Simplify your answers.)

b. Compute

projWy

and

UUTy.

projWy=

	409
	559
	359

and

UUTy=

	409
	559
	359

(Simplify your answers.)

6.3.17

Let

y=

	4
	8
	1

,

u1=

	−23
	23
	13

,

u2=

	23
	13
	23

,

and

W=Span u1,u2.

Complete parts (a) and (b).

a. Let

U=

	u1	u2

.

Compute

UTU

and

UUT.

UTU=

	1	0
	0	1

and

UUT=

	89	−29	29
	−29	59	49
	29	49	59

(Simplify your answers.)

b. Compute

projWy

and

UUTy.

projWy=

	2
	4
	5

and

UUTy=

	2
	4
	5

(Simplify your answers.)

6.3.19

Let

u1=

	−2
	−2
	2

,

u2=

	−2
	1
	−1

,

and

u3=

	0
	0
	1

.

Note that

u1

and

u2

are orthogonal but that

u3

is not orthogonal to

u1

or

u2.

It can be shown that

u3

is not in the subspace W spanned by

u1

and

u2.

Use this fact to construct a nonzero vector

v

in

ℝ3

that is orthogonal to

u1

and

u2.

A nonzero vector in

ℝ3

that is orthogonal to

u1

and

u2

is

v=

	0
	12
	12

.

6.3.19

Let

u1=

	−1
	1
	−1

,

u2=

	−2
	−1
	1

,

and

u3=

	0
	0
	1

.

Note that

u1

and

u2

are orthogonal but that

u3

is not orthogonal to

u1

or

u2.

It can be shown that

u3

is not in the subspace W spanned by

u1

and

u2.

Use this fact to construct a nonzero vector

v

in

ℝ3

that is orthogonal to

u1

and

u2.

A nonzero vector in

ℝ3

that is orthogonal to

u1

and

u2

is

v=

	0
	12
	12

.

6.3.19

Let

u1=

	2
	1
	1

,

u2=

	−3
	3
	3

,

and

u3=

	0
	0
	1

.

Note that

u1

and

u2

are orthogonal but that

u3

is not orthogonal to

u1

or

u2.

It can be shown that

u3

is not in the subspace W spanned by

u1

and

u2.

Use this fact to construct a nonzero vector

v

in

ℝ3

that is orthogonal to

u1

and

u2.

A nonzero vector in

ℝ3

that is orthogonal to

u1

and

u2

is

v=

	0
	−12
	12

.

6.3.21

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

If

z

is orthogonal to

u1

and

u2

and if

W=Spanu1, u2,

then

z

must be in

W⊥.

Choose the correct answer below.

The statement is false. Since

z

is orthogonal to

u1

and

u2,

it exists in

Spanu1, u2.

Since

W=Spanu1, u2,

z

is in W and cannot be in

W⊥.

The statement is false. If

z

is orthogonal to

u1

and

u2,

it only follows that

z

is orthogonal to

Spanu1

and

Spanu2.

This is not enough information to conclude that

z

is in

W⊥.

The statement is true. Since

z

is orthogonal to

u1

and

u2,

it is orthogonal to every vector in

Spanu1, u2,

a set that spans W.

The statement is true.

W⊥

is the set of all vectors orthogonal to

u1

and

u2,

so by definition,

z

is in

W⊥.

6.3.21

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

If

z

is orthogonal to

u1

and

u2

and if

W=Spanu1, u2,

then

z

must be in

W⊥.

Choose the correct answer below.

The statement is false. Since

z

is orthogonal to

u1

and

u2,

it exists in

Spanu1, u2.

Since

W=Spanu1, u2,

z

is in W and cannot be in

W⊥.

The statement is true. Since

z

is orthogonal to

u1

and

u2,

it is orthogonal to every vector in

Spanu1, u2,

a set that spans W.

The statement is false. If

z

is orthogonal to

u1

and

u2,

it only follows that

z

is orthogonal to

Spanu1

and

Spanu2.

This is not enough information to conclude that

z

is in

W⊥.

The statement is true.

W⊥

is the set of all vectors orthogonal to

u1

and

u2,

so by definition,

z

is in

W⊥.

6.3.22

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

For each

y

and each subspace W, the vector

y−projWy

is orthogonal to W.

Choose the correct answer below.

The statement is false. Because

y

can be written uniquely in the form

y=projWy+z,

where

z

is in W and

projWy

is in

W⊥,

it follows that

z=y−projWy.

The statement is true. Because

y

can be written uniquely in the form

y=projWy+z,

where

projWy

is in W and

z

is in

W⊥,

it follows that

z=y−projWy.

The statement is true. Because

y

and

projWy

are both orthogonal to W, a linear combination of them must also be orthogonal to W.

The statement is false. Because

y−projWy

is in W, it cannot be orthogonal to W.

6.3.22

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

For each

y

and each subspace W, the vector

y−projWy

is orthogonal to W.

Choose the correct answer below.

The statement is true. Because

y

and

projWy

are both orthogonal to W, a linear combination of them must also be orthogonal to W.

The statement is true. Because

y

can be written uniquely in the form

y=projWy+z,

where

projWy

is in W and

z

is in

W⊥,

it follows that

z=y−projWy.

The statement is false. Because

y−projWy

is in W, it cannot be orthogonal to W.

The statement is false. Because

y

can be written uniquely in the form

y=projWy+z,

where

z

is in W and

projWy

is in

W⊥,

it follows that

z=y−projWy.

6.3.23

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

The orthogonal projection

y

of

y

onto a subspace W can sometimes depend on the orthogonal basis for W used to compute

y.

Choose the correct answer below.

The statement is false. The uniqueness property of the orthogonal decomposition

y=y+z

indicates that, no matter the basis used to find it, the decomposition will always be the same.

The statement is false. The orthogonal projection

y

of

y

onto a subspace W depends on an orthonormal basis for W.

The statement is true. For each possible orthogonal basis of W,

y

is expressed as a different linear combination of the vectors in that basis.

The statement is true. The orthogonal projection

y

of

y

onto a subspace W depends on an orthonormal basis for W.

6.3.24

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

If

y

is in a subspace W, then the orthogonal projection of

y

onto W is

y

itself.

Choose the correct answer below.

The statement is true. For an orthogonal basis

B=u1, ..., up

of W,

y

and

projWy

can be written as linear combinations of vectors in B with equal weights.

The statement is true. If

y

is in W, then

projWy=−y,

which is in the same spanning set as

y.

The statement is false. If

y

is in W, then

projWy=0.

This means the statement is false unless

y=0.

The statement is false. If

y

is in W, then

projWy

is orthogonal to

y

and is in

W⊥.

6.3.24

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

If

y

is in a subspace W, then the orthogonal projection of

y

onto W is

y

itself.

Choose the correct answer below.

The statement is false. If

y

is in W, then

projWy=0.

This means the statement is false unless

y=0.

The statement is true. For an orthogonal basis

B=u1, ..., up

of W,

y

and

projWy

can be written as linear combinations of vectors in B with equal weights.

The statement is false. If

y

is in W, then

projWy

is orthogonal to

y

and is in

W⊥.

The statement is true. If

y

is in W, then

projWy=−y,

which is in the same spanning set as

y.

6.3.25

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

The best approximation to

y

by elements of a subspace W is given by the vector

y−projWy.

The statement is

false.

The

Best Approximation Theorem

says that the best approximation to

y

is

projWy.

6.3.25

Determine whether the statement below is true or false. Justify the answer. Assume all vectors and subspaces are in

ℝn.

The best approximation to

y

by elements of a subspace W is given by the vector

y−projWy.

The statement is

false.

The

Best Approximation Theorem

says that the best approximation to

y

is

projWy.

6.4.1

The given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

0 8 8 ,  5 6 −2

An orthogonal basis for W is

	0
	8
	8

,

	5
	4
	−4

.

(Type a vector or list of vectors. Use a comma to separate vectors as needed.)

6.4.1

The given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

0 8 8 ,  8 8 −4

An orthogonal basis for W is

	0
	8
	8

,

	8
	6
	−6

.

(Type a vector or list of vectors. Use a comma to separate vectors as needed.)

6.4.10

Find an orthogonal basis for the column space of the matrix to the right.

−1 5 6 3 −6 2 2 −1 6 1 −5 −2

An orthogonal basis for the column space of the given matrix is

	−1
	3
	2
	1

,

	3
	0
	3
	−3

,

	2
	0
	0
	2

.

(Type a vector or list of vectors. Use a comma to separate vectors as needed.)

6.4.10

Find an orthogonal basis for the column space of the matrix to the right.

−1 6 5 2 −8 4 1 −2 7 1 −4 −3

An orthogonal basis for the column space of the given matrix is

	−1
	2
	1
	1

,

	2
	0
	2
	0

,

	0
	2
	0
	−4

.

(Type a vector or list of vectors. Use a comma to separate vectors as needed.)

6.4.10

Find an orthogonal basis for the column space of the matrix to the right.

−1 6 6 1 −8 4 1 −2 6 1 −4 −2

An orthogonal basis for the column space of the given matrix is

	−1
	1
	1
	1

,

	1
	−3
	3
	1

,

	6
	5
	4
	−3

.

(Type a vector or list of vectors. Use a comma to separate vectors as needed.)

6.4.14

The columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that A=QR.

A= −2 −3 5 7 −2 −2 −4 −1 , Q= −27 −114 57 214 −27 0 −47 314

R=

	7	7
	0	14

6.4.14

The columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that A=QR.

A= 2 −3 5 7 −2 −2 −4 −4 , Q= 27 −529 57 229 −27 0 −47 0

R=

	7	7
	0	29

6.4.15

An orthogonal basis for the column space of matrix A is v1, v2, v3. Use this orthogonal basis to find a QR factorization of matrix A.

A= 1 2 4 −1 −2 1 0 4 2 1 6 4 1 6 7 ,   v1= 1 −1 0 1 1 ,   v2= −2 2 4 2 2 ,   v3= 2 3 −1 −1 2

Q=

	12	−232	219
	−12	232	319
	0	432	−119
	12	232	−119
	12	232	219

,

R=

	2	8	7
	0	42	32
	0	0	19

(Type exact answers, using radicals as needed.)

6.4.17

Determine whether the following statement is true or false, where all the vectors and subspaces are in

ℝn.

Justify the answer.

If

v1, v2, v3

is an orthogonal basis for

W,

then multiplying

v3

by a scalar c gives a new orthogonal basis

v1, v2, cv3.

Choose the correct answer below.

The statement is false. If the scale factor is negative it will not give a new orthogonal basis.

The statement is true. For example, when the vectors in an orthogonal set are normalized, they are multiplied by scalars, and the set of new vectors is orthogonal.

The statement is true. Multiplication by a scalar, c, will change the length of

v3,

but it will not change the relative angles between the vectors and so they will still be orthogonal.

The statement is false. If the scale factor is zero it will not give a new orthogonal basis.

6.4.17

Determine whether the following statement is true or false, where all the vectors and subspaces are in

ℝn.

Justify the answer.

If

v1, v2, v3

is an orthogonal basis for

W,

then multiplying

v3

by a scalar c gives a new orthogonal basis

v1, v2, cv3.

Choose the correct answer below.

The statement is false. If the scale factor is zero it will not give a new orthogonal basis.

The statement is false. If the scale factor is negative it will not give a new orthogonal basis.

The statement is true. For example, when the vectors in an orthogonal set are normalized, they are multiplied by scalars, and the set of new vectors is orthogonal.

The statement is true. Multiplication by a scalar, c, will change the length of

v3,

but it will not change the relative angles between the vectors and so they will still be orthogonal.

6.4.21

Determine whether the following statement is true or false, where all the vectors and subspaces are in

ℝn.

Justify the answer.

If

A=QR,

where Q has orthonormal columns, then

R=QTA.

Choose the correct answer below.

The statement is false. If Q is square, then

QT=Q−1

and

R=Q−1A.

However, if Q is not square, the matrix

Q−1

does not exist and this statement is false.

The statement is false. Since Q has orthonormal columns,

QTQ=0.

So

QTA=QT(QR)=0.

The statement is true. Since Q has orthonormal columns,

QTQ=1.

So

QTA=QT(QR)=1R=R.

The statement is true. Since Q has orthonormal columns,

QTQ=I.

So

QTA=QT(QR)=IR=R.

6.4.21

Determine whether the following statement is true or false, where all the vectors and subspaces are in

ℝn.

Justify the answer.

If

A=QR,

where Q has orthonormal columns, then

R=QTA.

Choose the correct answer below.

The statement is false. If Q is square, then

QT=Q−1

and

R=Q−1A.

However, if Q is not square, the matrix

Q−1

does not exist and this statement is false.

The statement is true. Since Q has orthonormal columns,

QTQ=1.

So

QTA=QT(QR)=1R=R.

The statement is true. Since Q has orthonormal columns,

QTQ=I.

So

QTA=QT(QR)=IR=R.

The statement is false. Since Q has orthonormal columns,

QTQ=0.

So

QTA=QT(QR)=0.

6.5.1

Find a least-squares solution of

Ax=b

by (a) constructing the normal equations for

x

and (b) solving for

x.

A=

	−1	2
	2	−3
	−1	1

,

b=

	2
	1
	3

a. Construct the normal equations for

x.

	6	−9
	−9	14

	x1
	x2

=

	−3
	4

(Simplify your answers.)

b. Solve for

x.

x=

	−2
	−1

(Simplify your answer.)

6.5.1

Find a least-squares solution of

Ax=b

by (a) constructing the normal equations for

x

and (b) solving for

x.

A=

	−1	1
	1	−2
	−1	1

,

b=

	1
	2
	3

a. Construct the normal equations for

x.

	3	−4
	−4	6

	x1
	x2

=

	−2
	0

(Simplify your answers.)

b. Solve for

x.

x=

	−6
	−4

(Simplify your answer.)

6.5.3

Find a least-squares solution of

Ax=b

by (a) constructing the normal equations for

x

and (b) solving for

x.

A=

	1	−4
	−1	4
	0	2
	4	7

,

b=

	5
	1
	−3
	4

a. Construct the normal equations for

x

without solving.

	18	20
	20	85

x=

	20
	6

(Simplify your answers.)

b. Solve for

x.

x=

	158113
	−146565

(Simplify your answer.)

6.5.4

Find a least-squares solution of

Ax=b

by (a) constructing the normal equations for

x

and (b) solving for

x.

A=

	1	1
	1	−1
	1	1

,

b=

	4
	6
	0

a. Construct the normal equations for

x.

	3	1
	1	3

	x1
	x2

=

	10
	−2

(Simplify your answers.)

b. Solve for

x.

x=

	4
	−2

(Simplify your answer.)

6.5.5

Describe all least-squares solutions of the equation

Ax=b.

A=

	1	1	0
	1	0	1
	1	0	1
	1	1	0

,

b=

	3
	11
	3
	1

The least-squares solutions are all vectors of the form

x=

	7
	−5
	0

+x3

	−1
	1
	1

,

with

x3

free.

(Simplify your answers.)

6.5.5

Describe all least-squares solutions of the equation

Ax=b.

A=

	1	0	1
	1	0	1
	1	1	0
	1	1	0

,

b=

	2
	10
	1
	3

The least-squares solutions are all vectors of the form

x=

	6
	−4
	0

+x3

	−1
	1
	1

,

with

x3

free.

(Simplify your answers.)

6.5.7

Compute the least-squares error associated with the least-squares solution

x

of

Ax=b.

A=

	1	−2
	−1	2
	0	4
	5	9

,

b=

	4
	1
	−2
	4

,

x=

	1,5131,154
	−3491,154

The least-squares error is

155771154.

(Type an exact answer, using radicals as needed.)

6.5.8

For

A=

1	2
1	−1
1	1

and

b=

3

8

0

,

a least-squares solution of

Ax=b

is

x=

5

−2

.

Compute the least-squares error associated with this solution.

The least-squares error is

14.

(Simplify your answer. Type an exact answer, using radicals as needed.)

6.5.10

Find (a) the orthogonal projection of b onto Col A and (b) a least-squares solution of

Ax=b.

A=

−1	4
1	8
−1	4

,

b=

11

−1

3

a. The orthogonal projection of b onto Col A is

b=

	7
	−1
	7

.

(Simplify your answers.)

b. A least-squares solution of

Ax=b

is

x=

	−5
	12

.

(Simplify your answers.)

6.5.13

Let

A=

	9	7
	−5	1
	9	7

,

b=

	10
	−7
	6

,

u=

	9
	−5

,

and

v=

	9
	−3

.

Compute

Au

and

Av,

and compare them with

b.

Could

u

possibly be a least-squares solution of

Ax=b?

(Answer this without computing a least-squares solution.)

Au=

	46
	−50
	46

(Simplify your answer.)

Av=

	60
	−48
	60

(Simplify your answer.)

Compare

Au

and

Av

with

b.

Could

u

possibly be a least-squares solution of

Ax=b?

Au

is closer to

b

than

Av

is. Thus,

u

could

possibly be a least-squares solution of

Ax=b.

Av

is closer to

b

than

Au

is. Thus,

u

cannot

possibly be a least-squares solution of

Ax=b.

Au

and

Av

are equally close to

b.

Thus, both

u

and

v

can be a least-squares solution of

Ax=b.

Au

and

Av

are equally close to

b.

Thus, neither

u

nor

v

can be a least-squares solution of

Ax=b.

6.5.13

Let

A=

	8	9
	−6	1
	8	9

,

b=

	11
	−5
	9

,

u=

	5
	−3

,

and

v=

	5
	−6

.

Compute

Au

and

Av,

and compare them with

b.

Could

u

possibly be a least-squares solution of

Ax=b?

(Answer this without computing a least-squares solution.)

Au=

	13
	−33
	13

(Simplify your answer.)

Av=

	−14
	−36
	−14

(Simplify your answer.)

Compare

Au

and

Av

with

b.

Could

u

possibly be a least-squares solution of

Ax=b?

Au

is closer to

b

than

Av

is. Thus,

u

could

possibly be a least-squares solution of

Ax=b.

Av

is closer to

b

than

Au

is. Thus,

u

cannot

possibly be a least-squares solution of

Ax=b.

Au

and

Av

are equally close to

b.

Thus, both

u

and

v

can be a least-squares solution of

Ax=b.

Au

and

Av

are equally close to

b.

Thus, neither

u

nor

v

can be a least-squares solution of

Ax=b.

6.5.15

Use the factorization

A=QR

to find the least-squares solution of

Ax=b.

A=

	2	3
	2	4
	1	1

=

	23	−13
	23	23
	13	−23

	3	5
	0	1

b=

	7
	3
	1

x=

	4
	−1

(Simplify your answer.)

6.5.17

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

The general least-squares problem is to find an

x

that makes

Ax

as close as possible to

b.

Choose the correct answer below.

The statement is false. The general least-squares problem attempts to find an

x

such that

Ax=b.

The statement is false. The general least-squares problem attempts to find an

x

that maximizes

b−Ax.

The statement is true. The general least-squares problem attempts to find an

x

that minimizes

b−Ax.

The statement is true. The general least-squares problem attempts to find an

x

such that

Ax=b.

6.5.18

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

If

b

is in the column space of A, then every solution of

Ax=b

is a least-squares solution.

The statement is

true.

A least-squares solution,

x,

is a vector that makes

Ax

the closest point in

Col A

to

b.

Since

b

is in

Col A,

all solutions of

Ax=b

are

least-squares solutions.

6.5.19

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

A least-squares solution of

Ax=b

is a vector

x

that satisfies

Ax=b,

where

b

is the orthogonal projection of

b

onto

Col A.

Choose the correct answer below.

The statement is false. Since

b

is not the closest point in

Col A

to

b,

Ax=b

is not necessarily consistent.

The statement is false. The equations

Ax=b

and

Ax=b

are not equivalent.

The statement is true. The equations

Ax=b

and

Ax=b

are equivalent.

The statement is true. Since

b

is the closest point in

Col A

to

b,

Ax=b

is consistent and

x

is such that

Ax=b

is a least-squares solution of

Ax=b.

6.5.20

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

A least-squares solution of

Ax=b

is a vector

x

such that

b−Ax≤b−Ax

for all

x

in

ℝn.

Choose the correct answer below.

The statement is true because the general least-squares problem attempts to find an

x

that maximizes

b−Ax.

The statement is false because a least-squares solution of

Ax=b

is a vector

x

such that

b−Ax<b−Ax

for all

x

in

ℝn.

The statement is false because a least-squares solution of

Ax=b

is a vector

x

such that

b−Ax≤b−Ax

for all

x

in

ℝn.

The statement is true because the general least-squares problem attempts to find an

x

that minimizes

b−Ax.

6.5.21

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

Any solution of

ATAx=ATb

is a least-squares solution of

Ax=b.

Choose the correct answer below.

The statement is true. The equation

ATAx=ATb

is equivalent to

Ax=b.

The statement is false. The set of least-squares solutions of

Ax=b

coincides with the nonempty set of solutions of the normal equations, defined as

ATx=b.

The statement is false. The set of least-squares solutions of

Ax=b

coincides with the nonempty set of solutions of the normal equations, defined as

ATx=ATAb.

The statement is true. The set of least-squares solutions of

Ax=b

coincides with the nonempty set of solutions of the normal equations, defined as

ATAx=ATb.

6.5.22

Determine whether the statement below is true or false. Justify the answer. Assume A is an

m×n

matrix and

b

is in

ℝm.

If the columns of A are linearly independent, then the equation

Ax=b

has exactly one least-squares solution.

Choose the correct answer below.

The statement is true. If the columns of A are linearly independent, then

ATA

is invertible and

x=ATA−1ATb

is the least-squares solution to

Ax=b.

The statement is false. If the columns of A are linearly independent, then there are infinitely many least-squares solutions to

Ax=b.

The statement is true. If the columns of A are linearly independent, then

AAT

is invertible and

x=AAT−1ATb

is the least-squares solution to

Ax=b.

The statement is false. The least-squares solution is only guaranteed to be unique if the columns of A are orthogonal.

6.6.1

Find the equation

y=β0+β1x

of the least-squares line that best fits the given data points.

(1,2),

(2,2),

(3,4),

(4,4)

The line is

y=1+45x.

(Type integers or decimals.)

6.6.3

Find the equation

y=β0+β1x

of the least-squares line that best fits the given data points.

(−1,0),

(0,1),

(1,2),

(2,4)

The line is

y=1110+1310x.

(Type integers or decimals.)

6.6.9

If you enter the following data into a machine and it returns a y value of

19

when

x=2.5,

should you trust the machine? Justify your answer.

(1,2),

(2,2),

(3,3),

(4,3)

Choose the correct answer below.

Yes, the machine should be trusted. Machine learning models are not limited to straight lines. A fifth or greater degree polynomial could intersect all four of the original data points and still pass through the predicted point at

(2.5,

19).

No, the machine should not be trusted. Since the data are not collinear, a linear model will not make any sense. A prediction like

y=19,

which is so far from the range of y values in the data is the kind of nonsensical prediction that results from using poor data.

Yes, the machine should be trusted. Although the predicted value of

y=19

may seem surprisingly large, machines very rarely make calculation errors.

No, the machine should not be trusted.

x=2.5

is within the range of x values in the data, while

y=19

is significantly greater than the maximum y value in the data. It is unreasonable to expect any model to predict a value so far outside the range of y values

6.6.13

A certain experiment produces the data

(1, 1.7),

(2, 2.8),

(3, 3.3),

(4, 3.6),

and

(5, 3.9).

Describe the model that produces a least-squares fit of these points by a function of the form

y=β1x+β2x2.

Such a function might arise, for example, as the revenue from the sale of x units of a product, when the amount offered for sale affects the price to be set for the product. Answer parts (a) through (c) below.

a. Give the design matrix, the observation vector, and the unknown parameter vector. Choose the correct design matrix X below.

	1	1
	2	4
	3	9
	4	16
	5	25

	1	1	1	1	1
	1	2	3	4	5
	1	4	9	16	25

	1	1	1
	1	2	4
	1	3	9
	1	4	16
	1	5	25

	1	2	3	4	5
	1	4	9	16	25

The observation vector is

y=

	1.7
	2.8
	3.3
	3.6
	3.9

.

Choose the correct form of the parameter vector

β

below.

	β1
	β2
	β3
	β4
	β5

0

β1

β2

x

x2

	β1
	β2

b. Find the associated least-squares curve for the data.

y=1.72x+−0.19x2

(Round to two decimal places as needed.)

c. If a machine learned the curve you found in (b), what output would it provide for an input of

x=6?

y=3.48

(Round to two decimal places as needed.)

*9.1.B1

Determine the following sums or differences.

(a)

(9+5i)+(2+7i)

                          (b)

(5+8i)−(−7−4i)

(a)

(9+5i)+(2+7i)=11+12i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(5+8i)−(−7−4i)=12+12i

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B1

Determine the following sums or differences.

(a)

(5−7i)+(3−9i)

                          (b)

(4+2i)−(−7−5i)

(a)

(5−7i)+(3−9i)=8−16i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(4+2i)−(−7−5i)=11+7i

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B1

Determine the following sums or differences.

(a)

(1−7i)+(6−9i)

                          (b)

(5−8i)−(−7+6i)

(a)

(1−7i)+(6−9i)=7−16i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(5−8i)−(−7+6i)=12−14i

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B2

Express the following products in standard form.

(a)

(−3+4i)(3−9i)

                             (b)

(−5−i)(5−i)

(a)

(−3+4i)(3−9i)=27+39i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(−5−i)(5−i)=−26+0i

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B2

Express the following products in standard form.

(a)

(−6+2i)(3−7i)

                             (b)

(−3−i)(3−i)

(a)

(−6+2i)(3−7i)=−4+48i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(−3−i)(3−i)=−10+0i

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B2

Express the following products in standard form.

(a)

(−4+3i)(8−4i)

                             (b)

(−3−i)(3−i)

(a)

(−4+3i)(8−4i)=−20+40i

(Simplify your answer. Type your answer in the form

a+bi.)

(b)

(−3−i)(3−i)=−10

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B3

Determine the complex conjugates of the following numbers.

(a)	17i	(b)	3
(c)	11−12i	(d)	14+13i

(a) The complex conjugate of

17i

is

0−17i.

(Simplify your answer. Type your answer in the form

a+bi.)

(b) The complex conjugate of

3

is

3+0i.

(Simplify your answer. Type your answer in the form

a+bi.)

(c) The complex conjugate of

11−12i

is

11+12i.

(Simplify your answer. Type your answer in the form

a+bi.)

(d) The complex conjugate of

14+13i

is

14−13i.

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B3

Determine the complex conjugates of the following numbers.

(a)	18i	(b)	10
(c)	6−4i	(d)	3+5i

(a) The complex conjugate of

18i

is

0−18i.

(Simplify your answer. Type your answer in the form

a+bi.)

(b) The complex conjugate of

10

is

10+0i.

(Simplify your answer. Type your answer in the form

a+bi.)

(c) The complex conjugate of

6−4i

is

6+4i.

(Simplify your answer. Type your answer in the form

a+bi.)

(d) The complex conjugate of

3+5i

is

3−5i.

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B3

Determine the complex conjugates of the following numbers.

(a)	19i	(b)	4
(c)	16−11i	(d)	15+5i

(a) The complex conjugate of

19i

is

−19i.

(Simplify your answer. Type your answer in the form

a+bi.)

(b) The complex conjugate of

4

is

4.

(Simplify your answer. Type your answer in the form

a+bi.)

(c) The complex conjugate of

16−11i

is

16+11i.

(Simplify your answer. Type your answer in the form

a+bi.)

(d) The complex conjugate of

15+5i

is

15−5i.

(Simplify your answer. Type your answer in the form

a+bi.)

*9.1.B5

Express the following quotients in standard form.

(a)

34−8i

(b)

1−6i4+8i

(a)

34−8i=320+310i

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

(b)

1−6i4+8i=−1120−25i

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

*9.1.B5

Express the following quotients in standard form.

(a)

67−8i

(b)

5−9i7+8i

(a)

67−8i=42113+48i113

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

(b)

5−9i7+8i=−37113−103113i

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

*9.1.B5

Express the following quotients in standard form.

(a)

95−8i

(b)

4−6i5+8i

(a)

95−8i=4589+7289i

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

(b)

4−6i5+8i=−2889−6289i

(Simplify your answer. Type your answer in the form

a+bi.

Use integers or fractions for any numbers in the expression.)

*9.1.B6

Use polar form to determine

z1z2

and

z1z2.

z1=−1−3i,

z2=3−i         

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=18cos14π+isin14π

z1z2=4cos−56π+isin−56π

z1z2=4cos−12π+isin−12π

z1z2=cos−56π+isin−56π

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=322cos14π+isin14π

z1z2=cos−12π+isin−12π

z1z2=cos−56π+isin−56π

z1z2=4cos−12π+isin−12π

*9.1.B6

Use polar form to determine

z1z2

and

z1z2.

z1=1−i,

z2=−3−3i         

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=6cos12π+isin12π

z1z2=62cos14π+isin14π

z1z2=13(cos(−π)+isin(−π))

z1z2=6(cos(−π)+isin(−π))

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=13(cos(−π)+isin(−π))

z1z2=cos14π+isin14π

z1z2=13cos12π+isin12π

z1z2=6cos12π+isin12π

*9.1.B6

Use polar form to determine

z1z2

and

z1z2.

z1=−1−3i,

z2=−3−3i         

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=62cos112π+isin112π

z1z2=62cos−1712π+isin−1712π

z1z2=4cos−13π+isin−13π

z1z2=232cos−1712π+isin−1712π

Use polar form to determine

z1z2.

Choose the correct answer below.

z1z2=232cos112π+isin112π

z1z2=62cos112π+isin112π

z1z2=232cos−1712π+isin−1712π

z1z2=cos−13π+isin−13π

*9.1.B7

Use polar form to determine the following.

(−3−3i)3

(−3−3i)3=54−54i

(Simplify your answer. Type your answer in the form

a+bi.

Type an exact answer, using radicals as needed.)

*9.1.B7

Use polar form to determine the following.

3+i4

3+i4=−8+83i

(Simplify your answer. Type your answer in the form

a+bi.

Type an exact answer, using radicals as needed.)

*9.1.B7

Use polar form to determine the following.

(−3−3i)5

(−3−3i)5=972+972i

(Simplify your answer. Type your answer in the form

a+bi.

Type an exact answer, using radicals as needed.)

*9.2.B1

For parts (a) and (b) below, determine whether the system is consistent, and if it is, determine the general solution.

(a)	z2−iz3	=	5
	iz1−z2+(−1−i)z3	=	−7−9i
	−2iz1+z2+(3+2i)z3	=	15+16i

Choose the correct answer below.

z=

	1+2i
	3+4i
	4+2i

z=

	1+2i
	0
	0

+s

	3+4i
	4+2i
	1

z=

	1+2i
	3+4i
	0

+s

	4+2i
	0
	1

The system is inconsistent.

(b)	z1+(−1−i)z2+iz3	=	−2−i
	iz1+(1−i)z2+3iz4	=	i
	z1+(−1−i)z2+(1+i)z3+3iz4	=	−3+2i

Choose the correct answer below.

z=

	1
	0
	−1+3i
	0

z=

	1
	0
	−1+3i
	0

+s

	1+i
	1
	0
	0

+t

	−3
	0
	−3i
	1

z=

	1
	0
	−1+3i
	0

+s

	1+i
	1
	0
	0

The system is inconsistent.

*9.3.B1

Calculate the following.

(a) −3+5i 5i − −2−2i −4−3i

(b) −3i −5−5i −5+4i

(c) (−3−2i) 3+4i −1−4i −4+i

(a)

	−3+5i
	5i

−

	−2−2i
	−4−3i

=

	−1+7i
	4+8i

(Simplify your answer.)

(b)

−3i

	−5−5i
	−5+4i

=

	−15+15i
	12+15i

(Simplify your answer.)

(c)

(−3−2i)

	3+4i
	−1−4i
	−4+i

=

	−1−18i
	−5+14i
	14+5i

(Simplify your answer.)

*9.3.B1

Calculate the following.

(a) 3−i 2−4i + −1+i 4−2i

(b) 3i 2−5i 1+2i

(c) (2+4i) 4−i −3−3i −3+2i

(a)

	3−i
	2−4i

+

	−1+i
	4−2i

=

	2
	6−6i

(Simplify your answer.)

(b)

3i

	2−5i
	1+2i

=

	15+6i
	−6+3i

(Simplify your answer.)

(c)

(2+4i)

	4−i
	−3−3i
	−3+2i

=

	12+14i
	6−18i
	−14−8i

(Simplify your answer.)

*9.3.B1

Calculate the following.

(a) −3i 1+5i − 4+5i 3+4i

(b) 4i −3+i −2+4i

(c) (−1+4i) −5i 2−5i −1−3i

(a)

	−3i
	1+5i

−

	4+5i
	3+4i

=

	−4−8i
	−2+i

(Simplify your answer.)

(b)

4i

	−3+i
	−2+4i

=

	−4−12i
	−16−8i

(Simplify your answer.)

(c)

(−1+4i)

	−5i
	2−5i
	−1−3i

=

	20+5i
	18+13i
	13−i

(Simplify your answer.)

*9.3.B3

Determine which of the following sets is a basis for

ℂ3.

(a)

	0
	−2i
	0

, 

	5
	0
	−2

, 

	0
	−2
	−i

     

(b)

	3i
	3i
	3i

, 

	0
	0
	i

, 

	−3
	−i
	1

(a) Is the given set a basis for

ℂ3?

No, because it is a linearly independent set that spans

ℂ3.

Yes, because it is a linearly dependent set that spans

ℂ3.

No, because one of the vectors in the set is not in

ℂ3.

Yes, because it contains three vectors in

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

No, because it is a linearly dependent set that does not span

ℂ3.

(b) Is the given set a basis for

ℂ3?

Yes, because it is a linearly dependent set that spans

ℂ3.

Yes, because it contains three vectors in

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

No, because it is a linearly independent set that spans

ℂ3.

No, because only one of the vectors in the set is in

ℂ3.

No, because it is a linearly dependent set that does not span

ℂ3.

*9.3.B3

Determine which of the following sets is a basis for

ℂ3.

(a)

	5
	1
	−2

, 

	0
	−2
	−2i

, 

	1
	−2i
	0

     

(b)

	i
	i
	i

, 

	−1
	−1
	i

, 

	−2
	−2i
	−1

(a) Is the given set a basis for

ℂ3?

No, because it is a linearly dependent set that does not span

ℂ3.

Yes, because it contains three vectors in

ℂ3.

No, because one of the vectors in the set is not in

ℂ3.

No, because it is a linearly independent set that spans

ℂ3.

Yes, because it is a linearly dependent set that spans

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

(b) Is the given set a basis for

ℂ3?

No, because it is a linearly dependent set that does not span

ℂ3.

No, because it is a linearly independent set that spans

ℂ3.

Yes, because it is a linearly dependent set that spans

ℂ3.

Yes, because it contains three vectors in

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

No, because only one of the vectors in the set is in

ℂ3.

*9.3.B3

Determine which of the following sets is a basis for

ℂ3.

(a)

	3
	2
	0

, 

	2
	−1
	2i

, 

	0
	i
	1

     

(b)

	0
	i
	1

, 

	0
	−2
	i

, 

	2i
	2i
	2i

(a) Is the given set a basis for

ℂ3?

No, because it is a linearly independent set that spans

ℂ3.

No, because one of the vectors in the set is not in

ℂ3.

Yes, because it contains three vectors in

ℂ3.

No, because it is a linearly dependent set that does not span

ℂ3.

Yes, because it is a linearly dependent set that spans

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

(b) Is the given set a basis for

ℂ3?

No, because it is a linearly independent set that spans

ℂ3.

No, because only one of the vectors in the set is in

ℂ3.

No, because it is a linearly dependent set that does not span

ℂ3.

Yes, because it contains three vectors in

ℂ3.

Yes, because it is a linearly independent set that spans

ℂ3.

Yes, because it is a linearly dependent set that spans

ℂ3.

*9.3.B4

Find a basis for the rowspace, columnspace, and nullspace of the following matrices.

(a)

A=

	2	2	2i
	2i	2i	2
	1+i	1+i	−1−i

(b)

B=

	0	i	2−2i
	1+7i	2i	5
	3	0	3i

(a) Determine a basis for Row(A). Choose the correct answer below.

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

	1
	1
	0

,

	0
	0
	1

	1
	0
	0

,

	0
	1
	1

	1
	2
	i

Determine a basis for Col(A). Choose the correct answer below.

	2i
	2
	−1−i

	2
	2i
	1+i

,

	2
	2i
	1+i

,

	2i
	2
	−1−i

	2
	2i
	1+i

,

	2i
	2
	−1−i

	2
	2i
	1+i

Determine a basis for Null(A). Choose the correct answer below.

	0
	−1
	1

	−1
	1
	0

	−2
	1
	0

,

	−i
	0
	1

The basis is the empty set.

(b) Determine a basis for Row(B). Choose the correct answer below.

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

	1
	11
	11i

	1
	1
	0

,

	0
	0
	1

	1
	0
	0

,

	0
	1
	5

Determine a basis for Col(B). Choose the correct answer below.

	0
	1+7i
	3

	0
	1+7i
	3

,

	i
	2i
	0

,

	2−2i
	5
	3i

	0
	1+7i
	3

,

	2−2i
	5
	3i

	0
	1+7i
	3

,

	i
	2i
	0

Determine a basis for Null(B). Choose the correct answer below.

	−11
	1
	0

,

	−11i
	0
	1

	−1
	1
	0

	0
	−1
	5

The basis is the empty set.

*9.3.B4

Find a basis for the rowspace, columnspace, and nullspace of the following matrices.

(a)

A=

	4	4	4i
	4i	4i	4
	2+2i	2+2i	−2−2i

                                            (b)

B=

	0	2i	4−3i
	2+6i	i	3
	2	0	3i

(a) Determine a basis for Row(A). Choose the correct answer below.

	1
	4
	2i

	1
	0
	0

,

	0
	1
	2

	1
	1
	0

,

	0
	0
	1

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

Determine a basis for Col(A). Choose the correct answer below.

	4
	4i
	2+2i

	4
	4i
	2+2i

,

	4i
	4
	−2−2i

	4
	4i
	2+2i

,

	4
	4i
	2+2i

,

	4i
	4
	−2−2i

	4i
	4
	−2−2i

Determine a basis for Null(A). Choose the correct answer below.

	−1
	1
	0

	−4
	1
	0

,

	−2i
	0
	1

	0
	−2
	1

The basis is the empty set.

(b) Determine a basis for Row(B). Choose the correct answer below.

	1
	2
	0

,

	0
	0
	1

	1
	11
	9i

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

	1
	0
	0

,

	0
	1
	3

Determine a basis for Col(B). Choose the correct answer below.

	0
	2+6i
	2

	0
	2+6i
	2

,

	2i
	i
	0

,

	4−3i
	3
	3i

	0
	2+6i
	2

,

	2i
	i
	0

	0
	2+6i
	2

,

	4−3i
	3
	3i

Determine a basis for Null(B). Choose the correct answer below.

	−1
	2
	0

	0
	−2
	3

	−11
	1
	0

,

	−9i
	0
	1

The basis is the empty set.

*9.3.B4

Find a basis for the rowspace, columnspace, and nullspace of the following matrices.

(a)

A=

	2	2	2i
	2i	2i	2
	1+i	1+i	−1−i

                 (b)

B=

	0	2i	3−3i
	3+3i	3i	3
	2	0	3i

(a) Determine a basis for Row(A). Choose the correct answer below.

	1
	0
	0

,

	0
	1
	1

	1
	2
	i

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

	1
	1
	0

,

	0
	0
	1

Determine a basis for Col(A). Choose the correct answer below.

	2
	2i
	1+i

,

	2i
	2
	−1−i

	2i
	2
	−1−i

	2
	2i
	1+i

	2
	2i
	1+i

,

	2
	2i
	1+i

,

	2i
	2
	−1−i

Determine a basis for Null(A). Choose the correct answer below.

	−1
	1
	0

	−2
	1
	0

,

	−i
	0
	1

	0
	−1
	1

The basis is the empty set.

(b) Determine a basis for Row(B). Choose the correct answer below.

	1
	11
	8i

	1
	0
	0

,

	0
	1
	0

,

	0
	0
	1

	1
	2
	0

,

	0
	0
	1

	1
	0
	0

,

	0
	1
	3

Determine a basis for Col(B). Choose the correct answer below.

	0
	3+3i
	2

,

	2i
	3i
	0

	0
	3+3i
	2

,

	2i
	3i
	0

,

	3−3i
	3
	3i

	0
	3+3i
	2

,

	3−3i
	3
	3i

	0
	3+3i
	2

Determine a basis for Null(B). Choose the correct answer below.

	0
	−1
	3

	−11
	1
	0

,

	−8i
	0
	1

	−1
	2
	0

The basis is the empty set.